NCERT Solutions for Class 12 Maths Chapter 7 Integrals 2021

NCERT Solutions for Class 12 Maths Chapter 7 Integrals

Edited By Ramraj Saini | Updated on Sep 14, 2023 08:18 PM IST | #CBSE Class 12th

NCERT Integrals Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 7 Integrals are discussed here. This chapter deals with definite and indefinite integrals. Integration class 12 also includes elementary properties of integration including basic techniques of integration. NCERT Class 12 maths chapter 7 solutions will be very helpful when you are solving the questions from NCERT books for Class 12 Maths. These NCERT Class 12 Maths solutions chapter 7 are prepared by subject matter experts that are very easy to understand. students can practice integrals class 12 ncert solutions to get good hold on the concepts.

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NCERT Integrals Class 12 Questions And Answers
NCERT Integrals Class 12 Questions And Answers PDF Free Download
Class 12 maths chapter 7 NCERT Solutions - Important Formulae
NCERT Integrals Class 12 Questions And Answers (Intext Questions and Exercise)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals - Topics
NCERT solutions for class 12 maths - Chapter wise
Key Features of NCERT Solutions for Class 12 Maths Chapter 7 Integrals
NCERT solutions for class 12 - subject wise
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 12 Maths Chapter 7 Integrals

NCERT solutions for class 12 math chapter 7 integrals are important for board exams as well as for competitive examinations like JEE Main, VITEEE, BITSAT, etc but without command, on the concepts of integrals ncert solutions meritorious marks cant be scoured. Therefore chapter 7 maths class 12 is recommended to students. Also, you can check the NCERT solutions for other Classes here.

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NCERT Integrals Class 12 Questions And Answers PDF Free Download

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Class 12 maths chapter 7 NCERT Solutions - Important Formulae

>> Integration as Inverse of Differentiation: Integration is the inverse process of differentiation.

In differential calculus, we find the derivative of a given function, while in integral calculus, we find a function whose derivative is given.

Indefinite Integrals:

∫f(x) dx = F(x) + C

These integrals are called indefinite integrals or general integrals.

C is an arbitrary constant that leads to different anti-derivatives of the given function.

Multiple Anti-Derivatives:

A derivative of a function is unique, but a function can have infinite anti-derivatives or integrals.

Properties of Indefinite Integral:

∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx

For any real number k, ∫k f(x) dx = k∫f(x)dx.

In general, if f1, f2, …, fn are functions and k1, k2, …, kn are real numbers, then ∫[k1f1(x) + k2f2(x) + … + knfn(x)] dx = k1 ∫f1(x) dx + k2 ∫f2(x) dx + … + kn ∫fn(x) dx

First Fundamental Theorem of Integral Calculus:

Define the area function A(x) = ∫[a, x]f(t)dt for x ≥ a, where f is continuous on [a, b].

Then A'(x) = f(x) for every x ∈ [a, b].

Second Fundamental Theorem of Integral Calculus:

If f is a continuous function on [a, b], then ∫[a, b]f(x)dx = F(b) - F(a), where F(x) is an antiderivative of f(x).

Standard Integral Formulas:

∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ -1)

∫cos x dx = sin x + C

∫sin x dx = -cos x + C

∫sec² x dx = tan x + C

∫cosec² x dx = -cot x + C

∫sec x tan x dx = sec x + C

∫cosec x cot x dx = -cosec x + C

∫e^x dx = e^x + C

∫a^x dx = (a^x)/ln(a) + C

∫(1/x) dx = ln|x| + C

Other Integral Formulas:

∫tan x dx = ln|sec x| + C

∫cot x dx = ln|sin x| + C

∫sec x dx = ln|sec x + tan x| + C

∫cosec x dx = ln|cosec x - cot x| + C

Free download Class 12 Maths Chapter 7 Question Answer for CBSE Exam.

NCERT Integrals Class 12 Questions And Answers (Intext Questions and Exercise)

Class 12 Integrals NCERT solutions Exercise: 7.1

Question:1 Find an anti derivative (or integral) of the following functions by the method of inspection. $\sin 2x$

Answer:

GIven $\sin 2x$ ;

So, the anti derivative of $\sin 2x$ is a function of x whose derivative is $\sin 2x$ .

$\frac{d}{dx}\left ( \cos 2x \right ) = -2\sin 2x$

$\implies \sin 2x =\frac{-1}{2} \frac{d}{dx}\left ( \cos 2x \right )$

Therefore, we have $\implies \sin 2x = \frac{d}{dx}\left ( \frac{-1}{2}\cos 2x \right )$

Or, antiderivative of $\sin 2x$ is $\left ( \frac{-1}{2}\cos 2x \right )$ .

Question:2 Find an anti derivative (or integral) of the following functions by the method of inspection. $\cos 3x$

Answer:

GIven $\cos 3x$ ;

So, the antiderivative of $\cos 3x$ is a function of x whose derivative is $\cos 3x$ .

$\frac{d}{dx}\left ( \sin 3x \right ) = 3\cos3x$

$\implies \cos 3x =\frac{1}{3} \frac{d}{dx}\left ( \sin 3x \right )$

Therefore, we have the anti derivative of $\cos 3x$ is $\frac{1}{3}\sin 3x$ .

Question:3 Find an anti derivative (or integral) of the following functions by the method of inspection. $e ^{2x}$

Answer:

GIven $e ^{2x}$ ;

So, the anti derivative of $e ^{2x}$ is a function of x whose derivative is $e ^{2x}$ .

$\frac{d}{dx}\left ( e ^{2x}\right ) = 2e ^{2x}$

$\implies e ^{2x} = \frac{1}{2}\frac{d}{dx}(e ^{2x})$

$\therefore e ^{2x} = \frac{d}{dx}(\frac{1}{2}e ^{2x})$

Therefore, we have the anti derivative of $e^{2x}$ is $\frac{1}{2}e ^{2x}$ .

Question:4 Find an anti derivative (or integral) of the following functions by the method of inspection. $( ax + b )^2$

Answer:

GIven $( ax + b )^2$ ;

So, the anti derivative of $( ax + b )^2$ is a function of x whose derivative is $( ax + b )^2$ .

$\frac{d}{dx} (ax+b)^3 = 3a(ax+b)^2$

$\Rightarrow (ax+b)^2 =\frac{1}{3a}\frac{d}{dx}(ax+b)^3$

$\therefore (ax+b)^2 = \frac{d}{dx}[\frac{1}{3a}(ax+b)^3]$

Therefore, we have the anti derivative of $(ax+b)^2$ is $[\frac{1}{3a}(ax+b)^3]$ .

Question:5 Find an anti derivative (or integral) of the following functions by the method of inspection. $\sin 2x - 4 e ^{3x}$

Answer:

GIven $\sin 2x - 4 e ^{3x}$ ;

So, the anti derivative of $\sin 2x - 4 e ^{3x}$ is a function of x whose derivative is $\sin 2x - 4 e ^{3x}$ .

$\frac{d}{dx} (-\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x}) = \sin 2x -4e^{3x}$

Therefore, we have the anti derivative of $\sin 2x - 4 e ^{3x}$ is $\left ( -\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x} \right )$ .

Question:6 Find the following integrals

$\int ( 4e ^{3x}+1) dx$

Answer:

Given intergral $\int ( 4e ^{3x}+1) dx$ ;

$4\int e ^{3x} dx + \int 1 dx = 4\left ( \frac{e^{3x}}{3} \right ) +x +C$

or $\frac{4}{3} e^{3x} +x +C$ , where C is any constant value.

Question:7 Find the following integrals $\int x ^2 ( 1- \frac{1}{x^2})dx$

Answer:

Given intergral $\int x ^2 ( 1- \frac{1}{x^2})dx$ ;

$\int x^2 dx - \int1dx$

or $\frac{x^3}{3} - x +C$ , where C is any constant value.

Question:8 Find the following integrals $\int ( ax ^2 + bx + c ) dx$

Answer:

Given intergral $\int ( ax ^2 + bx + c ) dx$ ;

$\int ax^2\ dx + \int bx\ dx + \int c\ dx$

$= a\int x^2\ dx + b\int x\ dx + c\int dx$

$= a\frac{x^3}{3}+b\frac{x^2}{2}+cx +C$

or $\frac{ax^3}{3}+\frac{bx^2}{2}+cx +C$ , where C is any constant value.

Question:9 Find the following integrals intergration of $\int \left ( 2x^2 + e ^x \right ) dx$

Answer:

Given intergral $\int \left ( 2x^2 + e ^x \right ) dx$ ;

$\int 2x^2\ dx + \int e^{x}\ dx$

$= 2\int x^2\ dx + \int e^{x}\ dx$

$= 2\frac{x^3}{3}+e^{x} +C$

or $\frac{2x^3}{3}+e^{x} +C$ , where C is any constant value.

Question:10 Find the following integrals $\int \left ( \sqrt x - \frac{1}{\sqrt x } \right ) ^2 dx$

Answer:

Given integral $\int \left ( \sqrt x - \frac{1}{\sqrt x } \right ) ^2 dx$ ;

or $\int (x+\frac{1}{x}-2)\ dx$

$= \int x\ dx + \int \frac{1}{x}\ dx -2\int dx$

$= \frac{x^2}{2} + \ln|x| -2x +C$ , where C is any constant value.

Question:11 Find the following integrals intergration of $\int \frac{x^3 + 5x^2 - 4}{x^2} dx$

Answer:

Given intergral $\int \frac{x^3 + 5x^2 - 4}{x^2} dx$ ;

or $\int \frac{x^3}{x^2}\ dx+\int \frac{5x^2}{x^2}\ dx -4\int \frac{1}{x^2}\ dx$

$\int x\ dx + 5\int1. dx - 4\int x^{-2}\ dx$

$= \frac{x^2}{2}+5x-4\left ( \frac{x^{-1}}{-1} \right )+C$

Or, $\frac{x^2}{2}+5x+\frac{4}{x}+C$ , where C is any constant value.

Question:12 Find the following integrals $\int \frac{x^3+ 3x +4 }{\sqrt x } dx$

Answer:

Given intergral $\int \frac{x^3+ 3x +4 }{\sqrt x } dx$ ;

or $\int \frac{x^3}{x^{\frac{1}{2}}}\ dx+\int \frac{3x}{x^{\frac{1}{2}}}\ dx +4\int \frac{1}{x^{\frac{1}{2}}}\ dx$

$= \int x^{\frac{5}{2}}\ dx + 3\int x^{\frac{1}{2}}\ dx +4\int x^{-\frac{1}{2}}\ dx$

$=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left ( x^{\frac{3}{2}} \right )}{\frac{3}{2}}+\frac{4\left ( x^{\frac{1}{2}} \right )}{\frac{1}{2}} +C$

Or, $= \frac{2}{7}x^{\frac{7}{2}} +2x^{\frac{3}{2}}+8\sqrt{x} +C$ , where C is any constant value.

Question:13 Find the following integrals intergration of $\int \frac{x^3 - x^2 + x -1 }{x-1 } dx$

Answer:

Given integral $\int \frac{x^3 - x^2 + x -1 }{x-1 } dx$

It can be written as

$= \int \frac{x^2(x-1)+(x+1)}{(x-1)} dx$

Taking $(x-1)$ common out

$= \int \frac{(x-1)(x^2+1)}{(x-1)} dx$

Now, cancelling out the term $(x-1)$ from both numerator and denominator.

$= \int (x^2+1)dx$

Splitting the terms inside the brackets

$=\int x^2dx + \int 1dx$

$= \frac{x^3}{3}+x+c$

Question:14 Find the following integrals $\int (1-x) \sqrt x dx$

Answer:

Given intergral $\int (1-x) \sqrt x dx$ ;

$\int \sqrt{x}\ dx - \int x\sqrt{x}\ dx$ or

$\int x^{\frac{1}{2}}\ dx - \int x^{\frac{3}{2}} \ dx$

$= \frac{x^\frac{3}{2}}{\frac{3}{2}} - \frac{x^{\frac{5}{2}}}{\frac{5}{2}} +C$

$= \frac{2}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}}+C$ , where C is any constant value.

Question:15 Find the following integrals $\int \sqrt x ( 3x^2 + 2x +3 )dx$

Answer:

Given intergral $\int \sqrt x ( 3x^2 + 2x +3 )dx$ ;

$= \int 3x^2\sqrt{x}\ dx + \int 2x\sqrt{x}\ dx + \int 3\sqrt {x}\ dx$ or $= 3\int x^{\frac{5}{2}}\ dx + 2\int x^{\frac{3}{2}} \ dx +3\int x^{\frac{1}{2}} \ dx$

$= 3\frac{x^\frac{7}{2}}{\frac{7}{2}} +2\frac{x^{\frac{5}{2}}}{\frac{5}{2}} +3\frac{x^{\frac{3}{2}}}{\frac{3}{2}} +C$

$= \frac{6}{7}x^{\frac{7}{2}} + \frac{4}{5}x^{\frac{5}{2}}+ 2x^{\frac{3}{2}}+C$ , where C is any constant value.

Question:16 Find the following integrals $\int ( 2x - 3 \cos x + e ^x ) dx$

Answer:

Given integral $\int ( 2x - 3 \cos x + e ^x ) dx$ ;

splitting the integral as the sum of three integrals

$\int 2x\ dx -3 \int \cos x\ dx +\int e^{x}\ dx$

$= 2 \frac{x^2}{2} - 3 \sin x + e^x+C$

$= x^2 - 3 \sin x + e^x+C$ , where C is any constant value.

Question:17 Find the following integrals $\int ( 2 x ^2 - 3 \sin x + 5 \sqrt x ) dx$

Answer:

Given integral $\int ( 2 x ^2 - 3 \sin x + 5 \sqrt x ) dx$ ;

$2\int x^2\ dx -3\int \sin x\ dx + 5\int \sqrt {x}\ dx$

$= 2 \frac{x^3}{3} - 3(-\cos x ) +5\left ( \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

$= \frac{2x^3}{3} +3\cos x +\frac{10}{3} x^{\frac{3}{2}}+C$ , where C is any constant value.

Question:18 Find the following integrals $\int \sec x ( \sec x + \tan x ) dx$

Answer:

Given integral $\int \sec x ( \sec x + \tan x ) dx$ ;

$\int (\sec^2x+ \sec x \tan x ) \ dx$

Using the integral of trigonometric functions

$= \int (sec^2 x )\ dx+ \int \sec x \tan x\ dx$

$= \tan x + \sec x +C$ , where C is any constant value.

Question:19 Find the following integrals intergration of $\int \frac{sec ^2 x }{cosec ^2 x } dx$

Answer:

Given integral $\int \frac{sec ^2 x }{cosec ^2 x } dx$ ;

$\int \frac{\frac{1}{\cos^2x}}{\frac{1}{\sin^2 x}}\ dx$

$= \int \frac{\sin^2 x }{\cos ^2 x } \ dx$

$=\int (\sec^2 x-1 )\ dx$

$=\int \sec^2 x\ dx-\int1 \ dx$

$= \tan x -x+C$ , where C is any constant value.

Question:20 Find the following integrals $\int \frac{2- 3 \sin x }{\cos ^ 2 x } dx$

Answer:

Given integral $\int \frac{2- 3 \sin x }{\cos ^ 2 x } dx$ ;

$\int \left ( \frac{2}{\cos^2x}-\frac{3\sin x }{\cos^2 x} \right )\ dx$

Using antiderivative of trigonometric functions

$= 2\tan x -3\sec x +C$ , where C is any constant value.

Question:21 Choose the correct answer
The anti derivative of $\left ( \sqrt x + 1/ \sqrt x \right )$ equals

$A) \frac{1}{3}x ^{1/3} + 2 x ^{1/2}+ C \\\\ B) \frac{2}{3}x ^{2/3} + \frac{1}{2}x ^{2}+ C \\\\ C ) \frac{2}{3}x ^{3/2} + 2 x ^{1/2}+ C\\\\ D) \frac{3}{2}x ^{3/2} + \frac{1}{2} x ^{1/2}+ C$

Answer:

Given to find the anti derivative or integral of $\left ( \sqrt x + 1/ \sqrt x \right )$ ;

$\int \left ( \sqrt x + 1/ \sqrt x \right )\ dx$

$\int x^{\frac{1}{2}}\ dx + \int x^{-\frac{1}{2}}\ dx$

$= \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + \frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C$

$= \frac{2}{3}x^{\frac{3}{2}} + 2x^{\frac{1}{2}} +C$ , where C is any constant value.

Hence the correct option is (C).

Question:22 Choose the correct answer The anti derivative of

If $\frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}$ such that f (2) = 0. Then f (x) is

$A ) x ^ 4 + \frac{1}{x^3} - \frac{129 }{8} \\\\ B ) x ^ 3 + \frac{1}{x^4} - \frac{129 }{8} \\\\ C ) x ^ 4 + \frac{1}{x^3} + \frac{129 }{8}\\\\ D) x ^ 3 + \frac{1}{x^4} - \frac{129 }{8}$

Answer:

Given that the anti derivative of $\frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}$

So, $\frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}$

$f(x) = \int 4 x ^3 - \frac{3}{x^4}\ dx$

$f(x) = 4\int x ^3 - 3\int {x^{-4}}\ dx$

$f(x) = 4\left ( \frac{x^4}{4} \right ) -3\left ( \frac{x^{-3}}{-3} \right )+C$

$f(x) = x^4+\frac{1}{x^3} +C$

Now, to find the constant C;

we will put the condition given, f (2) = 0

$f(2) = 2^4+\frac{1}{2^3} +C = 0$

$16+\frac{1}{8} +C = 0$

or $C = \frac{-129}{8}$

$\Rightarrow f(x) = x^4+\frac{1}{x^3}-\frac{129}{8}$

Therefore the correct answer is A .

Class 12 Integrals NCERT solutions Exercise: 7.2

Question:1 Integrate the functions $\frac{2x}{1+ x ^2}$

Answer:

Given to integrate $\frac{2x}{1+ x ^2}$ function,

Let us assume $1+x^2 =t$

we get, $2xdx = dt$

$\implies \int \frac{2x}{1+x^2} dx = \int \frac{1}{t} dt$

$= \log|t| +C$

$= \log|1+x^2| +C$ now back substituting the value of $t = 1+x^2$

as $(1+x^2)$ is positive we can write

$= \log(1+x^2) +C$

Question:2 Integrate the functions $\frac{( \log x )^2}{x}$

Answer:

Given to integrate $\frac{( \log x )^2}{x}$ function,

Let us assume $\log |x| = t$

we get, $\frac{1}{x}dx= dt$

$\implies \int \frac{\left ( \log|x| \right )^2}{x}\ dx = \int t^2dt$

$= \frac{t^3}{3}+C$

$= \frac{(\log|x|)^3}{3}+C$

Question:3 Integrate the functions $\frac{1}{x+ x \log x }$

Answer:

Given to integrate $\frac{1}{x+ x \log x }$ function,

Let us assume $1+\log x = t$

we get, $\frac{1}{x}dx= dt$

$\implies \int \frac{1}{x(1+\log x )} dx = \int \frac{1}{t} dt$

$= \log|t| +C$

$= \log |1+ \log x | +C$

Question:4 Integrate the functions $\sin x \sin ( \cos x )$

Answer:

Given to integrate $\sin x \sin ( \cos x )$ function,

Let us assume $\cos x =t$

we get, $-\sin x dx =dt$

$\implies \int \sin x \sin(\cos x)dx = -\int \sin t dt$

$= -\left ( -\cos t \right ) +C$

$= \cos t +C$

Back substituting the value of t we get,

$= \cos (\cos x ) +C$

Question:5 Integrate the functions $\sin ( ax + b ) \cos ( ax + b )$

Answer:

Given to integrate $\sin ( ax + b ) \cos ( ax + b )$ function,

$\sin ( ax + b ) \cos ( ax + b ) = \frac{2\sin ( ax + b ) \cos ( ax + b )}{2} = \frac{\sin 2(ax+b)}{2}$

Let us assume $2(ax+b) = t$

we get, $2adx =dt$

$\int \frac{\sin 2(ax+b)}{2} dx = \frac{1}{2}\int \frac{\sin t}{2a} dt$

$= \frac{1}{4a}[-cos t] +C$

Now, by back substituting the value of t,

$= \frac{-1}{4a}[cos 2(ax+b)] +C$

Question:6 Integrate the functions $\sqrt { ax + b }$

Answer:

Given to integrate $\sqrt { ax + b }$ function,

Let us assume $(ax+b) = t$

we get, $adx =dt$

$dx = \frac{1}{a}dt$

$\Rightarrow \int(ax+b)^{\frac{1}{2}} dx = \frac{1}{a}\int t^{\frac{1}{2}}dt$

Now, by back substituting the value of t,

$= \frac{1}{a}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

$= \frac{2(ax+b)^\frac{3}{2}}{3a} +C$

Question:7 Integrate the functions $x \sqrt { x +2 }$

Answer:

Given function $x \sqrt { x +2 }$ ,

$\int x\sqrt{x+2}$

Assume the $(x+2) = t$ 19634

$\therefore dx =dt$

$\Rightarrow \int x\sqrt{x+2} dx = \int (t-2) \sqrt{t} dt$

$= \int (t-2) \sqrt{t} dt$

$= \int \left ( t^{\frac{3}{2}}-2t^{\frac{1}{2}} \right )dt$

$= \int t^{\frac{3}{2}}dt -2\int t^{\frac{1}{2}}dt$

$= \frac{t^{\frac{5}{2}}}{\frac{5}{2}} -2\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

$= \frac{2}{5}t^{\frac{5}{2}} -\frac{4}{3}t^{\frac{3}{2}} +C$

Back substituting the value of t in the above equation.

or, $\frac{2}{5}(x+2)^{\frac{5}{2}}- \frac{4}{3}(x+2)^\frac{3}{2} +C$ , where C is any constant value.

Question:8 Integrate the functions $x \sqrt { 1+ 2 x^2 }$

Answer:

Given function $x \sqrt { 1+ 2 x^2 }$ ,

$\int x \sqrt { 1+ 2 x^2 }\ dx$

Assume the $1+2x^2= t$

$\therefore 4xdx =dt$

$\Rightarrow \int x\sqrt{1+2x^2}dx = \int \frac{\sqrt {t}}{4} dt$

Or $= \frac{1}{4}\int t^{\frac{1}{2}} dt = \frac{1}{4}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

$= \frac{1}{6}(1+2x^2)^{\frac{3}{2}} +C$ , where C is any constant value.

Question:9 Integrate the functions $( 4x +2 ) \sqrt { x ^ 2 + x + 1 }$

Answer:

Given function $( 4x +2 ) \sqrt { x ^ 2 + x + 1 }$ ,

$\int ( 4x +2 ) \sqrt { x ^ 2 + x + 1 } dx$

Assume the $1+x+x^2 = t$

$\therefore (2x+1)dx =dt$

$\Rightarrow \int (4x+2)\sqrt{1+x+x^2} dx$

$= \int 2\sqrt {t}dt = 2\int \sqrt{t}dt$

$= 2\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

Now, back substituting the value of t in the above equation,

$= \frac{4}{3}(1+x+x^2)^{\frac{3}{2}} +C$ , where C is any constant value.

Question:10 Integrate the functions $\frac{1}{x - \sqrt x }$

Answer:

Given function $\frac{1}{x - \sqrt x }$ ,

$\int \frac{1}{x - \sqrt x } dx$

Can be written in the form:

$\frac{1}{x - \sqrt x } = \frac{1}{\sqrt {x}(\sqrt{x}-1)}$

Assume the $(\sqrt{x}-1) =t$

$\therefore \frac{1}{2\sqrt{x}}dx =dt$

$\Rightarrow \int \frac{1}{\sqrt{x}(\sqrt{x}-1)}dx = \int \frac{2}{t}dt$

$= 2\log|t| +C$

$= 2\log|\sqrt{x}-1| +C$ , where C is any constant value.

Question:11 Integrate the functions $\frac{x }{ \sqrt{ x +4} }$ , x > 0

Answer:

Given function $\frac{x }{ \sqrt{ x +4} }$ ,

$\int \frac{x }{ \sqrt{ x +4} }dx$

Assume the $x+4 =t$ so, $x =t-4$

$\therefore dx=dt$

$\int \frac{x}{\sqrt{x+4}}dx = \int \frac{t-4}{\sqrt{t}}dt$

$\int t^\frac{1}{2}dt -4\int t^{\frac{-1}{2}}dt$

$= \frac{2}{3}t^{\frac{3}{2}} - 4\left ( 2t^{\frac{1}{2}} \right )+C$

$= \frac{2}{3}(x+4)^{\frac{3}{2}} -16(x+4)^{\frac{1}{2}}+C$

, where C is any constant value.

Question:12 Integrate the functions $( x ^3 - 1 ) ^{1/3} x ^ 5$

Answer:

Given function $( x ^3 - 1 ) ^{1/3} x ^ 5$ ,

$\int ( x ^3 - 1 ) ^{1/3} x ^ 5 dx$

Assume the $x^3-1 = t$

$\therefore 3x^2dx=dt$

$\implies \int(x^3-1)^{\frac{1}{3}} x^5 dx = \int (x^3-1)^{\frac{1}{3}}x^3.x^2dx$

$= \int t^{\frac{1}{3}}(t+1)\frac{dt}{3}$

$= \frac{1}{3} \int \left ( t^\frac{4}{3}+t^\frac{1}{3} \right )dt$

$= \frac{1}{3}\left [ \frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right ]+C$

$= \frac{1}{3}\left [ \frac{3}{7}t^{\frac{7}{3}}+\frac{3}{4}t^{\frac{4}{3}} \right ]+C$

$= \frac{1}{7}(x^3-1)^{\frac{7}{3}} + \frac{1}{4}(x^3-1)^{\frac{4}{3}} +C$ , where C is any constant value.

Question:13 Integrate the functions $\frac{x ^2 }{(2+3x^3)^3}$

Answer:

Given function $\frac{x ^2 }{(2+3x^3)^3}$ ,

$\int \frac{x ^2 }{(2+3x^3)^3} dx$

Assume the $2+3x^3 =t$

$\therefore 9x^2dx=dt$

$\implies \int\frac{x^2}{(2+3x^2)}dx = \frac{1}{9}\int \frac{dt}{t^3}$

$= \frac{1}{9}\left ( \frac{t^{-2}}{-2} \right ) +C$

$= \frac{-1}{18}\left ( \frac{1}{t^2} \right )+C$

$= \frac{-1}{18(2+3x^3)^2}+C$ , where C is any constant value.

Question:14 Integrate the functions $\frac{1}{x (\log x )^m} , x > 0 , m \neq 1$

Answer:

Given function $\frac{1}{x (\log x )^m} , x > 0 , m \neq 1$ ,

Assume the $\log x =t$

$\therefore \frac{1}{x}dx =dt$

$\implies \int\frac{1}{x(logx)^m}dx = \int\frac{dt}{t^m}$

$=\left ( \frac{t^{-m+1}}{1-m} \right ) +C$

$= \frac{(log x )^{1-m}}{(1-m)} +C$ , where C is any constant value.

Question:15 Integrate the functions $\frac{x}{9- 4 x ^2 }$

Answer:

Given function $\frac{x}{9- 4 x ^2 }$ ,

Assume the $9-4x^2 =t$

$\therefore -8xdx =dt$

$\implies \int\frac{x}{9-4x^2} = -\frac{1}{8}\int \frac{1}{t}dt$

$= \frac{-1}{8}\log|t| +C$

Now back substituting the value of t ;

$= \frac{-1}{8}\log|9-4x^2| +C$ , where C is any constant value.

Question:16 Integrate the functions $e ^{ 2 x +3 }$

Answer:

Given function $e ^{ 2 x +3 }$ ,

Assume the $2x+3 =t$

$\therefore 2dx =dt$

$\implies \int e^{2x+3} dx = \frac{1}{2}\int e^t dt$

$= \frac{1}{2}e^t +C$

Now back substituting the value of t ;

$= \frac{1}{2}e^{2x+3}+C$ , where C is any constant value.

Question:17 Integrate the functions $\frac{x }{e^{x^{2}}}$

Answer:

Given function $\frac{x }{e^{x^{2}}}$ ,

Assume the $x^2=t$

$\therefore 2xdx =dt$

$\implies \int \frac{x}{e^{x^2}}dx = \frac{1}{2}\int \frac{1}{e^t}dt$

$= \frac{1}{2}\int e^{-t} dt$

$= \frac{1}{2}\left ( \frac{e^{-t}}{-1} \right ) +C$

$= \frac{-1}{2}e^{-x^2} +C$

$= \frac{-1}{2e^{x^2} }+C$ , where C is any constant value.

Question:18 Integrate the functions $\frac{e ^{\tan ^{-1}x}}{1+ x^2 }$

Answer:

Given,

$\frac{e ^{\tan ^{-1}x}}{1+ x^2 }$

Let's do the following substitution

$\\ tan^{-1}x = t \\ \implies \frac{1}{1+x^2}dx = dt$

$\therefore \int \frac{e ^{\tan ^{-1}x}}{1+ x^2 }dx = \int e ^{t}dt = e^t + C$

$= e^{tan^{-1}x} + C$

Question:19 Integrate the functions $\frac{e ^{2x}-1}{e ^{2x}+1}$

Answer:

Given function $\frac{e ^{2x}-1}{e ^{2x}+1}$ ,

Simplifying it by dividing both numerator and denominator by $e^x$ , we obtain

$\frac{\frac{e^{2x}-1}{e^x}}{\frac{e^{2x}+1}{e^x}} = \frac{e^x-e^{-x}}{e^x+e^{-x}}$

Assume the $e^{x}+e^{-x} =t$

$\therefore (e^x-e^{-x})dx =dt$

$\implies \int \frac{e^{2x}-1}{e^{2x}+1}dx = \int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$

$= \int \frac{dt}{t}$

$= \log |t| +C$

Now, back substituting the value of t,

$= \log |e^x+e^{-x}| +C$ , where C is any constant value.

Question:20 Integrate the functions $\frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }}$

Answer:

Given function $\frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }}$ ,

Assume the $e^{2x}+e^{-2x} =t$

$\therefore (2e^{2x}-2e^{-2x})dx =dt$

$\implies \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx = \int \frac{dt}{2t}$

$= \frac{1}{2}\int \frac{1}{t}dt$

$= \frac{1}{2}\log|t| +C$

Now, back substituting the value of t,

$= \frac{1}{2}\log|e^{2x}+e^{-2x}| +C$ , where C is any constant value.

Question:21 Integrate the functions $\tan ^2 ( 2x-3 )$

Answer:

Given function $\tan ^2 ( 2x-3 )$ ,

Assume the $2x-3 =t$

$\therefore 2dx =dt$

$\implies \int \tan^2(2x-3) dx = \frac{\int \tan^2(t)}{2}dt$

$=\frac{1}{2}\int (\sec^2t -1) dt$ $\left [\because \tan^2t+1 = \sec^2 t \right ]$

$= \frac{1}{2}\left [ \tan t - t \right ] +C$

Now, back substituting the value of t,

$= \frac{1}{2}\left [ \tan(2x-3)-2x+3 \right ]+C$

or $\frac{1}{2} \tan(2x-3)-x+C$ , where C is any constant value.

Question:22 Integrate the functions $\sec ^2 ( 7- 4x )$

Answer:

Given function $\sec ^2 ( 7- 4x )$ ,

Assume the $7-4x=t$

$\therefore -4dx =dt$

$\implies \int \sec^2(7-4x)dx = \frac{-1}{4}\int \sec^2t dt$

$=-\frac{1}{4}(\tan t) +C$

Now, back substituted the value of t.

$=-\frac{1}{4}\tan(7-4x)+C$ , where C is any constant value.

Question:23 Integrate the functions $\frac{\sin ^{-1}x}{\sqrt { 1- x^2 }}$

Answer:

Given function $\frac{\sin ^{-1}x}{\sqrt { 1- x^2 }}$ ,

Assume the $\sin^{-1}x =t$

$\therefore \frac{1}{\sqrt{1-x^2}}dx = dt$

$\implies \int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx =\int t dt$

$= \frac{t^2}{2}+C$

Now, back substituted the value of t.

$= \frac{(\sin^{-1}x)^2}{2}+C$ , where C is any constant value.

Question:24 Integrate the functions $\frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x }$

Answer:

Given function $\frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x }$ ,

or simplified as $\frac{2 \cos x - 3\sin x }{2(3 \cos x + 2 \sin x) }$

Assume the $3\cos x +2\sin x =t$

$\therefore (-3\sin x + 2\cos x )dx =dt$

$\implies \int \frac{2\cos x - 3\sin x }{6\cos x +4\sin x }dx = \int \frac{dt}{2t}$

$= \frac{1}{2}\int \frac{dt}{t}$

$= \frac{1}{2}\log|t| +C$

Now, back substituted the value of t.

$= \frac{1}{2}\log|3\cos x +2\sin x| +C$ , where C is any constant value.

Question:25 Integrate the functions $\frac{1 }{ \cos ^2 x (1-\tan x )^2}$

Answer:

Given function $\frac{1 }{ \cos ^2 x (1-\tan x )^2}$ ,

or simplified as $\frac{1 }{ \cos ^2 x (1-\tan x )^2} = \frac{\sec^2x}{(1-\tan x)^2}$

Assume the $(1-\tan x)=t$

$\therefore -\sec^2xdx =dt$

$\implies \int \frac{\sec^2x}{(1-\tan x)^2}dx = \int\frac{-dt}{t^2}$

$= -\int t^{-2} dt$

$= \frac{1}{t} +C$

Now, back substituted the value of t.

$= \frac{1}{1-\tan x}+C$

where C is any constant value.

Question:26 Integrate the functions $\frac{\cos \sqrt x }{\sqrt x }$

Answer:

Given function $\frac{\cos \sqrt x }{\sqrt x }$ ,

Assume the $\sqrt x =t$

$\therefore \frac{1}{2\sqrt x}dx =dt$

$\implies \int \frac{\cos \sqrt{x}}{\sqrt{x}}dx = 2\int \cos t dt$

$= 2\sin t +C$

Now, back substituted the value of t.

$= 2\sin \sqrt{x}+C$ , where C is any constant value.

Question:27 Integrate the functions $\sqrt { \sin 2x } \cos 2x$

Answer:

Given function $\sqrt { \sin 2x } \cos 2x$ ,

Assume the $\sin 2x = t$

$\therefore 2\cos 2x dx =dt$

$\implies \int \sqrt{\sin 2x }\cos 2x dx = \frac{1}{2}\int \sqrt t dt$

$= \frac{1}{2}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right )+C$

$= \frac{1}{3}t^{\frac{3}{2}}+C$

Now, back substituted the value of t.

$= \frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+C$ , where C is any constant value.

Question:28 Integrate the functions $\frac{\cos x }{\sqrt { 1+ \sin x }}$

Answer:

Given function $\frac{\cos x }{\sqrt { 1+ \sin x }}$ ,

Assume the $1+\sin x =t$

$\therefore \cos x dx = dt$

$\implies \int \frac{\cos x }{\sqrt{1+\sin x}}dx = \int \frac{dt}{\sqrt t}$

$= \frac{t^{\frac{1}{2}}}{\frac{1}{2}} +C$

$= 2\sqrt t +C$

Now, back substituted the value of t.

$= 2{\sqrt{1+\sin x}} +C$ , where C is any constant value.

Question:29 Integrate the functions $\cot x \: log \sin x$

Answer:

Given function $\cot x \: log \sin x$ ,

Assume the $\log \sin x =t$

$\therefore \frac{1}{\sin x }.\cos x dx =dt$

$\cot x dx =dt$

$\implies \int \cot x \log \sin x dx =\int t dt$

$= \frac{t^2}{2}+C$

Now, back substituted the value of t.

$= \frac{1}{2}(\log \sin x )^2+C$ , where C is any constant value.

Question:30 Integrate the functions $\frac{\sin x }{1+ \cos x }$

Answer:

Given function $\frac{\sin x }{1+ \cos x }$ ,

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

$\implies \int \frac{\sin x}{1+\cos x}dx = \int -\frac{dt}{t}$

$= -\log|t| +C$

Now, back substituted the value of t.

$= -\log|1+\cos x | +C$ , where C is any constant value.

Question:31 Integrate the functions $\frac{\sin x }{( 1+ \cos x )^2}$

Answer:

Given function $\frac{\sin x }{( 1+ \cos x )^2}$ ,

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

$\implies \int \frac{\sin x}{(1+\cos x)^2}dx = \int -\frac{dt}{t^2}$

$= -\int t^{-2}dt$

$= \frac{1}{t}+C$

Now, back substituted the value of t.

$= \frac{1}{1+\cos x} +C$ , where C is any constant value.

Question:32 Integrate the functions $\frac{1}{1+ \cot x }$

Answer:

Given function $\frac{1}{1+ \cot x }$

Assume that $I = \int \frac{1}{1+ \cot x } dx$

Now solving the assumed integral;

$I = \int \frac{1}{1+ \frac{\cos x }{\sin x} } dx$

$= \int \frac{\sin x }{\sin x + \cos x } dx$

$= \frac{1}{2}\int \frac{2\sin x }{\sin x + \cos x } dx$

$= \frac{1}{2}\int \frac{(\sin x+ \cos x ) +(\sin x -\cos x ) }{(\sin x + \cos x) } dx$

$=\frac{1}{2}\int 1 dx + \frac{1}{2} \int \frac{\sin x -\cos x }{\sin x +\cos x } dx$

$=\frac{1}{2}(x) + \frac{1}{2} \int \frac{\sin x -\cos x }{\sin x +\cos x } dx$

Now, to solve further we will assume $\sin x + \cos x =t$

Or, $(\cos x -\sin x)dx =dt$

$\therefore I = \frac{x}{2}+ \frac{1}{2}\int \frac{-(dt)}{t}$

$= \frac{x}{2}- \frac{1}{2}\log|t| +C$

Now, back substituting the value of t,

$= \frac{x}{2}- \frac{1}{2}\log|\sin x + \cos x| +C$

Question:33 Integrate the functions $\frac{1}{1- \tan x }$

Answer:

Given function $\frac{1}{1- \tan x }$

Assume that $I = \int \frac{1}{1- \tan x } dx$

Now solving the assumed integral;

$I = \int \frac{1}{1-\frac{\sin x}{\cos x }} dx$

$= \int \frac{\cos x }{\cos x - \sin x } dx$

$= \frac{1}{2}\int \frac{2\cos x }{\cos x - \sin x } dx$

$= \frac{1}{2}\int \frac{(\cos x -\sin x ) +(\cos x +\sin x ) }{(\cos x - \sin x) } dx$

$=\frac{1}{2}\int 1 dx + \frac{1}{2} \int \frac{\cos x +\sin x }{\cos x -\sin x } dx$

$=\frac{1}{2}(x) + \frac{1}{2} \int \frac{\cos x +\sin x }{\cos x -\sin x } dx$

Now, to solve further we will assume $\cos x - \sin x =t$

Or, $(-\sin x-\cos x )dx =dt$

$\therefore I = \frac{x}{2}+ \frac{1}{2}\int \frac{-(dt)}{t}$

$= \frac{x}{2}- \frac{1}{2}\log|t| +C$

Now, back substituting the value of t,

$= \frac{x}{2}- \frac{1}{2}\log|\cos x - \sin x| +C$

Question:34 Integrate the functions $\frac{\sqrt { \tan x } }{\sin x \cos x }$

Answer:

Given function $\frac{\sqrt { \tan x } }{\sin x \cos x }$

Assume that $I = \int \frac{\sqrt { \tan x } }{\sin x \cos x }dx$

Now solving the assumed integral;

Multiplying numerator and denominator by $\cos x$ ;

$I = \int \frac{\sqrt{\tan x }\times\cos x}{\sin x \cos x\times \cos x}dx$

$= \int \frac{\sqrt{\tan x }}{\tan x \cos^2 x } dx$

$= \int \frac{\sec^2 x }{\sqrt{\tan x }}dx$

Now, to solve further we will assume $\tan x =t$

Or, $\sec^2{x}dx =dt$

$\therefore I =\int \frac{dt}{\sqrt t}$

$=2\sqrt t +C$

Now, back substituting the value of t,

$= 2\sqrt{\tan x } +C$

Question:35 Integrate the functions $\frac{( 1+ \log x )^2}{x}$

Answer:

Given function $\frac{( 1+ \log x )^2}{x}$

Assume that $1+\log x =t$

$\therefore \frac{1}{x}dx =dt$

$= \int \frac{(1+\log x )^2}{x}dx = \int t^2 dt$

$= \frac{t^3}{3}+C$

Now, back substituting the value of t,

$= \frac{(1+\log x )^3}{3}+C$

Question:36 Integrate the functions $\frac{( x+1)( x+ \log x )^2}{x }$

Answer:

Given function $\frac{( x+1)( x+ \log x )^2}{x }$

Simplifying to solve easier;

$\frac{( x+1)( x+ \log x )^2}{x } = \left ( \frac{x+1}{x} \right )\left ( x+\log x \right )^2$

$=\left ( 1+\frac{1}{x} \right )\left ( x+\log x \right )^2$

Assume that $x+\log x =t$

$\therefore \left ( 1+\frac{1}{x} \right )dx = dt$

$= \int \left ( 1+\frac{1}{x} \right )\left ( x+\log x \right )^2 dx = \int t^2 dt$

$= \frac{t^3}{3}+C$

Now, back substituting the value of t,

$= \frac{(x+\log x )^3}{3}+C$

Question:37 Integrate the functions $\frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }$

Answer:

Given function $\frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }$

Assume that $x^4 =t$

$\therefore 4x^3 dx =dt$

$\Rightarrow \int \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }dx = \frac{1}{4} \int \frac{\sin(\tan^{-1} t)}{1+t^2}dt$ ......................(1)

Now to solve further we take $\tan ^{-1} t = u$

$\therefore \frac{1}{1+t^2} dt =du$

So, from the equation (1), we will get

$\Rightarrow \int \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }dx =\frac{1}{4}\int \sin u\ du$

$= \frac{1}{4}(-\cos u) +C$

Now back substitute the value of u,

$= \frac{-1}{4}\cos (\tan^{-1} t) +C$

and then back substituting the value of t,

$= \frac{-1}{4}\cos (\tan^{-1} x^4) +C$

Question:38 Choose the correct answer $\int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx\: \: \: equals$

$(A) 10^x - x^{10} + C \\\\(B) 10^x + x^{10} + C\\\\ (C) (10^x - x^{10})^{-1} + C \\\\ (D) log (10^x + x^{10}) + C$

Answer:

Given integral $\int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx$

Taking the denominator $x^{10} +10^x = t$

Now differentiating both sides we get

$\therefore \left ( 10x^9+10^x\log_{e}10 \right )dx = dt$

$\implies \int \frac{10x^9+10^x\log_{e}10}{x^{10}+10^x} dx = \int \frac{dt}{t}$

$= \log t +C$

Back substituting the value of t,

$= \log (x^{10}+10^x) +C$

Therefore the correct answer is D.

Question:39 Choose the correct answer $\int \frac{dx }{\sin ^ 2 x \cos ^2 x }\: \: \: equals$

$(A) \tan x + \cot x + C \\\\ (B) \tan x - \cot x + C\\\\ (C) \tan x \cot x + C\\\\ (D) \tan x - \cot 2x + C$

Answer:

Given integral $\int \frac{dx }{\sin ^ 2 x \cos ^ 2x }$

$\int \frac{dx }{\sin ^ 2 x \cos ^ 2x } = \int \frac{1}{\sin ^2 x \cos ^2 x } dx$

$=\int \frac{\sin ^2 x +\cos^2 x }{\sin^2 x \cos^2 x}dx$ $\left ( \because \sin ^2 x +\cos^2 x =1 \right )$

$=\int \frac{\sin^2 x }{\sin^2 x \cos^2 x}dx + \int \frac{\cos^2 x}{\sin^2 x \cos^2 x}dx$

$=\int \sec^2 x dx + \int cosec^2 x dx$

$=\tan x -\cot x +C$

Therefore, the correct answer is B.

Class 12 Integrals NCERT Solutions Exercise: 7.3

Question:1 Find the integrals of the functions $\sin ^ 2 ( 2x+ 5 )$

Answer:

$\sin ^ 2 ( 2x+ 5 )$

using the trigonometric identity

$sin^2x=\frac{1-cos2x}{2}$

we can write the given question as

= $\frac{1-\cos 2(2x+5)}{2}$ $= \frac{1-\cos (4x+10)}{2}$
$\\=\int \frac{1-\cos (4x+10)}{2}dx\\ =\frac{1}{2}\int dx - \frac{1}{2}\int \cos(4x+10)dx\\ =\frac{x}{2}-\frac{1}{2}[\sin(4x+10)/4]\\ =\frac{x}{2}-\frac{\sin(4x+10)}{8}+C$

Question:2 Find the integrals of the functions $\sin 3x \cos 4x$

Answer:

Using identity $\sin A\cos B = 1/2[sin(A+B)+sin(A-B)]$

, therefore the given integral can be written as

$\int \sin 3x\cos 4x=\frac{1}{2}\int sin(3x+4x)+sin(3x-4x)\ dx$

$=\frac{1}{2}\int sin(7x)-sin(x)\ dx\\ =1/2[\int \sin (7x) dx-\int \sin x\ dx]\\ =\frac{1}{2}[(-1/7)\cos 7x+\cos x+ C]\\ = \frac{\cos x}{2}-\frac{\cos 7x}{14}+C$

Question:3 Find the integrals of the functions $\cos 2x \cos 4x \cos 6x$

Answer:

Using identity
$cosAcosB = \frac{1}{2}[cos(A+B)+cos(A-B)]$

$\int \cos 2x.\cos 4x.\cos 6x = \int \cos 2x. \frac{1}{2}[(\cos 10x)+\cos 2x]dx$

Again use the same identity mentioned in the first line

$\\= \frac{1}{2}\int (\cos 2x.\cos 10x+\cos 2x. \cos 2x)dx\\ =\frac{1}{2}\int\frac{1}{2}({\cos12x +\cos 8x})dx+\frac{1}{2}\int (\frac{1+\cos 4x}{2})dx\\ =\frac{\sin 12x}{48}+\frac{\sin 8x}{32}+\frac{\sin 4x}{16}+ x/4+C$

Question:4 Find the integrals of the functions $\sin ^ 3 ( 2x +1 )$

Answer:

$\int \sin^3(2x+1)dx = \int \sin^2(2x+1).\sin(2x+1)dx$

The integral can be written as

$= \int (1-\cos^2(2x+1)).\sin(2x+1)dx$
Let
$\\\cos (2x+1) =t\\ \sin (2x+1)dx = -dt/2$

$\\=\frac{-1}{2}\int (1-t^2)dt\\ =\frac{-1}{2}[t-t^3/3]\\ =\frac{t^3}{6}-\frac{t}{2}$

Now, replace the value of t, we get;

$=\frac{\cos^3(2x+1)}{6}-\frac{\cos(2x+1)}{2}+C$

Question:5 Find the integrals of the functions $\sin ^3 x \cos ^ 3 x$

Answer:

$I = \int \sin^3x.\cos^3x\ dx$

rewrite the integral as follows

$\\=\int cos^3x.sin^2x.\sin x\ dx\\ =\int cos^3x(1-\cos^2x)\sin x\ dx$
Let $\cos x = t \Rightarrow \sin x dx =-dt$

$\\=-\int t^3(1-t^2)dt\\ =-\int(t^3-t^5)dt\\ =-[\frac{t^4}{4}]+[\frac{t^6}{6}] +C\\ =\frac{\cos^6x}{6}-\frac{cos^4x}{4}+C$ ......(replace the value of t as $cos\ x$ )

Question:6 Find the integrals of the functions $\sin x \sin 2x \sin 3x$

Answer:

Using the formula
$sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

we can write the integral as follows

$\int \sin x.\sin 2x\sin 3x\ dx = \int \sin x\frac{1}{2}[\cos x-\cos 5x]dx$
$\\=\frac{1}{2} \int [\sin x.\cos x-\sin x.\cos 5x]dx\\ =\frac{1}{2}\int \frac{\sin 2x}{2}dx-\frac{1}{2}\int \sin x. \cos 5x\ dx\\ =-\frac{\cos 2x}{8}-\frac{1}{4}\int[\sin 6x -\sin 4x]\\ =-\frac{\cos 2x}{8}-\frac{1}{4}[\frac{-\cos 6x}{6}+\frac{\cos 4x}{4}]\\ =-\frac{\cos 2x}{8}+\frac{\cos 6x}{24}-\frac{\cos 4x}{16}+C$

Question:7 Find the integrals of the functions $\sin 4x \sin 8x$

Answer:

Using identity

$sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

we can write the following integral as

$\sin 4x \sin 8x$ =
$\\=\frac{1}{2}\int(\cos 4x - \cos 12x) dx\\ =\frac{1}{2} [\int\cos 4x\ dx - \int \cos 12x\ dx]\\ =\frac{\sin 4x}{8}-\frac{\sin 12x}{24}+C$

Question:8 Find the integrals of the functions $\frac{1- \cos x }{1+ \cos x }$

Answer:

We know the identities

$\\1+\cos 2A = 2\cos^2A\\ 1-\cos 2A = 2\sin^2 A$

Using the above relations we can write

$\frac{1-\cos x}{1+\cos x}=\frac{\sin^2x/2}{\cos^2x/2} = \tan^2x/2$

$=\int \tan^2x/2 =\int (\sec^2x/2-1)dx$
$\\=\int (\sec^2x/2)dx-\int dx\\ = 2[\tan x/2]-{x}+C$

Question:9 Find the integrals of the functions $\frac {\cos x }{1 + \cos x }$

Answer:

The integral is rewritten using trigonometric identities

$\frac{\cos x}{1+ \cos x}= \frac{\cos^2x/2-\sin^2x/2}{2\cos^2x/2} =\frac{1}{2}[1-\tan^2x/2]$
$\\=\int \frac{1}{2}[1-\tan^2x/2] dx\\ =\frac{1}{2}\int 1-[sec^2\frac{x}{2}-1]=\frac{1}{2}\int 2-sec^2\frac{x}{2}\\=x-tan\frac{x}{2}+c$

Question:10 Find the integrals of the functions $\sin ^ 4 x$

Answer:

$\sin ^ 4 x$ can be written as follows using trigonometric identities
$\\=\sin^2x.\sin^2x\\ =\frac{1}{4}(1-\cos 2x)^{2}\\ =\frac{1}{4}(1+\cos^22x-2\cos 2x)\\ =\frac{1}{4}(1+\frac{1}{2}(1+\cos 4x)-2\cos 2x)\\ =3/8+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}$

Therefore,
$\Rightarrow \int \sin^4x\ dx = \int \frac{3}{8}dx+\frac{1}{8}\int\cos 4x\ dx -\frac{1}{2}\int\cos 2x\ dx$
$= \frac{3x}{8}+\frac{\sin 4x}{32} -\frac{\sin 2x}{4}+C$

Question:11 Find the integrals of the functions $\cos ^ 4 2x$

Answer:

$cos^42x=cos^32xcos2x$

now using the identity

$cos^3x=\frac{cos3x+3cosx}{4}$

$cos^32xcos2x=\frac{cos6x +3cos2x}{4}cos2x=\frac{cos6xcos2x+3cos^22x}{4}$

now using the below two identities

$cosacosb=\frac{cos(a+b)+cos(a-b)}{2}\\and\ cos^22x=\frac{1+cos4x}{2}\\$

the value

$cos^42x=cos^32xcos2x\\=\frac{cos6xcos2x+3cos^22x}{4}=\frac{cos 4x+cos8x}{8}+\frac{3}{4}\frac{1+cos4x}{2}$ .

the integral of the given function can be written as

$\int cos^42x=\int \frac{cos 4x+cos8x}{8}+\int \frac{3}{4}\frac{1+cos4x}{2}\\ \\=\frac{3}{8}x+\frac{sin4x}{8}+\frac{sin8x}{64}+C$

Question:12 Find the integrals of the functions $\frac{\sin ^ 2x }{1+ \cos x }$

Answer:

Using trigonometric identities we can write the given integral as follows.

$\frac{\sin ^ 2x }{1+ \cos x }$

$\\=\frac{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}{2\cos^2\frac{x}{2}}\\ =2\sin^2\frac{x}{2}\\ =1-\cos x$

$\therefore \int \frac{sin^22x}{1+\cos x} = \int (1-\cos x)dx$
$\\= \int 1dx-\int\cos x\ dx\\ =x-\sin x+C$

Question:13 Find the integrals of the functions $\frac{\cos 2x - \cos 2 \alpha }{\cos x - \cos \alpha }$

Answer:

We know that,

$\cos A-\cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$

Using this identity we can rewrite the given integral as

$\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos\alpha}=\frac{-2\sin\frac{2x+2\alpha}{2}\sin\frac{2x-2\alpha}{2}}{-2\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}$

$\\=\frac{\sin(x+\alpha)\sin(x-\alpha)}{\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}\\ =\frac{[2\sin\frac{x+\alpha}{2}\cos \frac{x+\alpha}{2}][2\sin\frac{x-\alpha}{2}\cos\frac{x-\alpha}{2}]}{\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}\\ =4\cos\frac{x+\alpha}{2}\cos\frac{x-\alpha}{2}\\ =2[\cos x+\cos \alpha]$

$\therefore \int\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos\alpha}=\int 2\cos x\ dx +\int 2\cos \alpha\ dx$
$=2[\sin x + x\cos \alpha]+C$

Question:14 Find the integrals of the functions $\frac{\cos x - \sin x }{1+ \sin 2x }$

Answer:

$\frac{\cos x-\sin x}{1+2\sin x}=\frac{\cos x-\sin x}{(sin^2x+cos^2x)+2 sin x.\cos x}$
$=\frac{\cos x-\sin x}{(\sin x+\cos x)^2}$

$\\sin x+\cos x =t\\ \therefore (\cos x-\sin x)dx = dt$

Now,
$=\int \frac{dt}{t^2}\\ =\int t^-2\ dt\\ =-t^{-1}+C\\ =-\frac{1}{(\sin x+\cos x)}+C$

Question:15 Find the integrals of the functions $\tan ^ 3 2x \sec 2x$

Answer:

$\tan^32x.\sec 2x = \tan^22x.\tan 2x.\sec 2x$
$\\= (\sec^22x-1).\tan 2x.\sec 2x\\ =\sec^22x.\tan 2x-\tan 2x.\sec 2x$

Therefore integration of $\tan ^ 3 2x \sec 2x$ =
$\\=\int\sec^22x.\tan 2x\ dx-\int\tan 2x.\sec 2x\ dx\\ =\int\sec^22x.\tan 2x\ dx-\sec 2x/2+C\\$ .....................(i)
Let assume

$\sec 2x = t$
So, that $2\sec 2x.\tan 2x\ dx =dt$
Now, the equation (i) becomes,

$\\\Rightarrow \frac{1}{2}\int t^2\ dt-\frac{\sec 2x}{2}+C\\ \Rightarrow \frac{t^3}{6}-\frac{\sec 2x}{2}+C\\ =\frac{(\sec 2x)^3}{6}-\frac{\sec 2x}{2}+C$

Question:16 Find the integrals of the functions $\tan ^ 4x$

Answer:

the given question can be rearranged using trigonometric identities

$tan^4x=(\sec^2x-1).\tan^2x\\ =\sec^2x.\tan^2x-\tan^2x\\ =\sec^2x.\tan^2x-\sec^2x+1$

Therefore, the integration of $\tan^4x$ = $\\=\int \sec^2x.\tan^2x\ dx-\int\sec^2x\ dx+\int dx\\ =(\int \sec^2x.\tan^2x\ dx)-\tan x+x+C\\$ ...................(i)
Considering only $\int \sec^2x.\tan^2x\ dx$
let $\tan x =t\Rightarrow \sec^2x\ dx =dt$

$\int \sec^2x\tan^2x\ dx = \int t^2\ dt = t^3/3=\frac{\tan^3x}{3}$

now the final solution is,

$\int \tan^4x =\frac{\tan^3x}{3}-\tan x+x+C$

Question:17 Find the integrals of the functions $\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

Answer:

$\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

now splitting the terms we can write

$\\=\frac{\sin^3x}{\sin^2x.\cos^2x}+\frac{\cos^3x}{\sin^2x.\cos^2x}\\ =\frac{\sin x}{cos^2x}+\frac{\cos x}{\sin^2x}\\ =\tan x.\sec x+\cot xcosec x$

Therefore, the integration of
$\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

$\\=\int (\tan x\sec x+\cot xcosec x)dx\\ =\sec x-cosec\ x+C$

Question:18 Find the integrals of the functions $\frac{\cos 2 x + 2 \sin ^ 2x }{\cos ^ 2 x }$

Answer:

The integral of the above equation is

$\\=\int (\frac{\cos 2x+2\sin^2x}{\cos^2x})dx\\ =\int (\frac{\cos 2x+(1-\cos 2x)}{\cos^2x}\\ =\int\frac{1}{\cos^2x}\\ =\int \sec^2x\ dx =\tan x+C$

Thus after evaluation, the value of integral is tanx+ c

Question:19 Find the integrals of the functions $\frac{1}{\sin x \cos ^3 x }$

Answer:

Let
We can write 1 = $\sin^2x +\cos^2x$
Then, the equation can be written as
$I =\frac{\sin^2x +\cos^2x}{\sin x\cos^3x}$

$I =\int (\tan x+\frac{1}{\tan x})\sec^2 x dx$
put the value of tan $x$ = t
So, that $\sec^2xdx =dt$

$\\\Rightarrow I=\int (t+\frac{1}{t})dt\\ =\frac{t^2}{2}+\log\left | t \right |+C\\ =\log\left | \tan x \right |+\frac{1}{2}\tan^2x+C$

Question:20 Find the integrals of the functions $\frac{\cos 2x }{( \cos x + \sin x )^2}$

Answer:

we know that $cos2x= cos^2x-sin^2x$
therefore,

$\frac{\cos 2x }{( \cos x + \sin x )^2}$
$\frac{\cos 2x}{1+\sin 2x}\\ \Rightarrow \int \frac{\cos 2x}{1+\sin 2x}\\$ let $1+sin2x =t \Rightarrow 2cos2x\ dx = dt$
Now the given integral can be written as

$\therefore \int \frac{\cos 2x}{(\cos x+\sin x)^2}=\frac{1}{2}\int \frac{1}{t}dt$
$\\\Rightarrow \frac{1}{2}\log\left | t \right |+C\\ \Rightarrow \frac{1}{2}\log\left | 1+\sin 2x \right |+C\\=log|sin^2x+cos^2x+2sinxcosx|+C\\=\frac{1}{2}log|(sinx+cosx)^2|+C=log|sinx+cosx|+C$

Question:21 Find the integrals of the functions $\sin ^ { -1} ( \cos x )$

Answer:

using the trigonometric identities we can evaluate the following integral as follows

$\dpi{100} \int \sin^{-1}(\cos x)dx = \int \sin^{-1}(sin(\frac{\pi}{2}-x))dx\\=\int(\frac{\pi}{2}-x)dx=\frac{\pi x}{2}-\frac{x^2}{2}+C$

Question:22 Find the integrals of the functions $\frac{1}{\cos ( x-a ) \cos ( x-b )}$

Answer:

Using the trigonometric identities following integrals can be simplified as follows

$\frac{1}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}[\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}]$

$=\frac{1}{\sin(a-b)}[\frac{\sin[(x-b)-(x-a)]}{\cos(x-a)\cos(x-b)}]$

$=\frac{1}{\sin(a-b)}[\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}]$

$=\frac{tan(x-b)-\tan (x-a)}{\sin(a-b)}$

$=\frac{1}{\sin(a-b)}\int tan(x-b)-\tan (x-a)dx$
$\\=\frac{1}{\sin(a-b)}[-\log\left | \cos(x-b) \right |+\log\left | \cos(x-a) \right |]\\ =\frac{1}{\sin(a-b)}(\log\left | \frac{\cos(x-a)}{\cos(x-b)} \right |)$

Question:23 Choose the correct answer

$\int \frac{\sin ^ 2 x - \cos ^ 2 x dx }{\sin ^ 2 x \cos ^ 2x } dx \: \:is \: \:equal \: \: to$ stgdrffd

Answer:

The correct option is (A)

On reducing the above integral becomes $\sec^2x-csc^2x$
$\int\sec^2x-csc^2x\ dx = \tan x+ \cot x+C$

Question:24 Choose the correct answer $\int \frac{e ^x ( 1+x)}{\cos ^ 2 ( e ^xx )} dx \: \: equals$

$\\(A) -\cot (ex^x) + C \\\\ (B) \tan (xe^x) + C\\\\ (C) \tan (e^x) + C \\\\ (D) \cot (e^x) + C$

Answer:

The correct option is (B)

Let $e^xx = t$ .
So, $(e^x.x+ 1.e^x)dx = dt$
(1+ $x$ ) $e^x\ dx =dt$

therefore,

$\int \frac{e^x(1+x)}{\cos^2(e^x.x)}dx =\int\frac{dt}{\cos^2t}$
$\\=\int \sec^2t dt\\ =\tan t +C\\ =\tan(e^x.x)+C$

NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.4

Question:1 Integrate the functions $\frac{3x^ 2 }{x^6 + 1 }$

Answer:

The given integral can be calculated as follows

Let $x^3 = t$
, therefore, $3x^2 dx =dt$

$\Rightarrow \int\frac{3x^2}{x^6+1}=\int \frac{dt}{t^2+1}$

$\\=\tan^{-1} t +C\\ =tan^{-1}(x^3)+C$

Question:2 Integrate the functions $\frac{1}{\sqrt { 1+ 4 x^2 }}$

Answer:

$\frac{1}{\sqrt { 1+ 4 x^2 }}$
let suppose 2x = t
therefore 2dx = dt

$\int \frac{1}{\sqrt{1+4x^2}} =\frac{1}{2}\int \frac{dt}{1+t^2}$
$\\=\frac{1}{2}[\log\left | t+\sqrt{1+t^2} \right |]+C\\ =\frac{1}{2}\log\left | 2x+\sqrt{4x^2+1} \right |+C$ .................using formula $\int\frac{1}{\sqrt{x^2+a^2}}dt = \log\left | x+\sqrt{x^2+a^2} \right |$

Question:3 Integrate the functions $\frac{1}{\sqrt { ( 2- x)^2+ 1 }}$

Answer:

$\frac{1}{\sqrt { ( 2- x)^2+ 1 }}$

let suppose 2-x =t
then, -dx =dt
$\Rightarrow \int\frac{1}{\sqrt{(2-x)^2+1}}dx = -\int \frac{1}{\sqrt{t^2+1}}dt$

using the identity

$\int \frac{1}{\sqrt{x^2+1}}dt=log\left | x+\sqrt{x^2+1} \right |$

$\\= -\log\left | t+\sqrt{t^2+1} \right |+C\\ =-\log\left | 2-x+\sqrt{(2-x)^2+1} \right |+C\\ =\log \left | \frac{1}{(2-x)+\sqrt{x^2-4x+5}} \right |+C$

Question:4 Integrate the functions $\frac{1}{\sqrt {9 - 25 x^2 }}$

Answer:

$\frac{1}{\sqrt {9 - 25 x^2 }}$
Let assume 5x =t,
then 5dx = dt

$\Rightarrow \int \frac{1}{\sqrt{9-25x^2}}=\frac{1}{5}\int \frac{1}{\sqrt{9-t^2}}dt$
$\\=\frac{1}{5}\int \frac{1}{\sqrt{3^2-t^2}}dt\\ =\frac{1}{5}\sin^{-1}(\frac{t}{3})+C\\ =\frac{1}{5}\sin^{-1}(\frac{5x}{3})+C$

The above result is obtained using the identity

$\\\int \frac{1}{\sqrt{a^2-x^2}}dt\\ =\frac{1}{a}sin^{-1}\frac{x}{a}$

Question:5 Integrate the functions $\frac{3x }{1+ 2 x ^ 4 }$

Answer:

$\frac{3x }{1+ 2 x ^ 4 }$

Let ${\sqrt{2}}x^2 = t$
$\therefore$ $2\sqrt{2}xdx =dt$

The integration can be done as follows

$\Rightarrow \int \frac{3x}{1+2x^4}= \frac{3}{2\sqrt{2}}\int \frac{dt}{1+t^2}$
$\\= \frac{3}{2\sqrt{2}}[\tan^{-1}t]+C\\ =\frac{3}{2\sqrt{2}}[\tan^{-1}(\sqrt{2}x^2)]+C$

Question:6 Integrate the functions $\frac{x ^ 2 }{1- x ^ 6 }$

Answer:

$\frac{x ^ 2 }{1- x ^ 6 }$

let $x^3 =t$
then $3x^2dx =dt$

using the special identities we can simplify the integral as follows

$\int \frac{x^2}{1-x^6}dx =\frac{1}{3}\int \frac{dt}{1-t^2}$
$=\frac{1}{3}[\frac{1}{2}\log\left | \frac{1+t}{1-t} \right |]+C\\ =\frac{1}{6}\log\left | \frac{1+x^3}{1-x^3} \right |+C$

Question:7 Integrate the functions $\frac{x-1 }{\sqrt { x^2 -1 }}$

Answer:

We can write above eq as
$\frac{x-1 }{\sqrt { x^2 -1 }}$ $=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx$ ............................................(i)

for $\int \frac{x}{\sqrt{x^2-1}}dx$ let $x^2-1 = t \Rightarrow 2xdx =dt$

$\therefore \int \frac{x}{\sqrt{x^2-1}}dx=\frac{1}{2}\int \frac{dt}{\sqrt{t}}$
$\\=\frac{1}{2}\int t^{1/2}dt\\ =\frac{1}{2}[2t^{1/2}]\\ =\sqrt{t}\\ =\sqrt{x^2-1}$
Now, by using eq (i)
$=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx$
$\\=\sqrt{x^2-1}-\int \frac{1}{\sqrt{x^2}-1}dx\\ =\sqrt{x^2-1}-\log\left | x+\sqrt{x^2-1} \right |+C$

Question:8 Integrate the functions $\frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}$

Answer:

The integration can be down as follows

$\frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}$
let $x^3 = t \Rightarrow 3x^2dx =dt$

$\therefore \frac{x^2}{\sqrt{x^6+a^6}}=\frac{1}{3}\int \frac{dt}{\sqrt{t^2+(a^3)^2}}$
$\\=\frac{1}{3}\log\left | t+\sqrt{t^2+a^6} \right |+C\\ =\frac{1}{3}\log\left | x^3+\sqrt{x^6+a^6} \right |+C$ ........................using $\int \frac{dx}{\sqrt{x^2+a^2}} = \log\left | x+\sqrt{x^2+a^2} \right |$

Question:9 Integrate the functions $\frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x+ 4 }}$

Answer:

The integral can be evaluated as follows

$\frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x + 4 }}$
let $\tan x =t \Rightarrow sec^2x dx =dt$

$\Rightarrow \int \frac{\sec^2x}{\sqrt{\tan^2x+4}}dx = \int \frac{dt}{\sqrt{t^2+2^2}}$
$\\= \log\left | t+\sqrt{t^2+4} \right |+C\\ =\log \left | \tan x+\sqrt{ tan^2x+4} \right |+C$

Question:10 Integrate the functions $\frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}$

Answer:

$\frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}$
the above equation can be also written as,
$=\int\frac{1}{\sqrt{(1+x)^2+1^2}}dx$
let 1+x = t
then dx = dt
therefore,

$\\=\int\frac{1}{\sqrt{t^2+1^2}}dx\\ =\log\left | t+\sqrt{t^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(1+x)^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(x^2+2x+2} \right |+C$

Question:11 Integrate the functions $\frac{1}{9 x ^2 + 6x + 5 }$

Answer:

$\frac{1}{9 x ^2 + 6x + 5 }$
this denominator can be written as
$9x^2+6x+5=9[x^2+\frac{2}{3}x+\frac{5}{9}]\\=9[(x+\frac{1}{3})^2+(\frac{2}{3})^2]$ Now,
$\frac{1}{9}\int \frac{1}{(x+\frac{1}{3})^2+(\frac{2}{3})^2}dx =\frac{1}{9} [\frac{3}{2}\tan^{-1}(\frac{(x+1/3)}{2/3})] +C\\=\frac{1}{6} \tan^{-1}(\frac{3x+1}{2})] +C$
......................................by using the form $(\int \frac{1}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a}))$

Question:12 Integrate the functions $\frac{1}{\sqrt{ 7-6x - x ^ 2 }}$

Answer:

the denominator can be also written as,
$7-6x-x^2=16-(x^2+6x+9)$
$=4^2-(x+3)^2$

therefore

$\int \frac{1}{\sqrt{7-6x-x^2}}dx=\int \frac{1}{\sqrt{4^2-(x+3)^2}}dx$
Let x+3 = t
then dx =dt

$\Rightarrow \int \frac{1}{\sqrt{4^2-(x+3)^2}}dx=\int \frac{1}{\sqrt{4^2-t^2}}dt$ ......................................using formula $\int \frac{1}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})$
$\\= sin^{-1}(\frac{t}{4})+C\\ =\sin^{-1}(\frac{x+3}{4})+C$

Question:13 Integrate the functions $\frac{1}{\sqrt { ( x-1)( x-2 )}}$

Answer:

(x-1)(x-2) can be also written as
= $x^2-3x+2$
= $(x-\frac{3}{2})^2-(\frac{1}{2})^2$

therefore

$\int \frac{1}{\sqrt{(x-1)(x-2)}}dx= \int \frac{1}{\sqrt{(x-\frac{3}{2})^2-(\frac{1}{2})^2}}dx$
let suppose
$x-3/2 = t \Rightarrow dx =dt$
Now,

$\Rightarrow \int \frac{1}{\sqrt{(x-\frac{3}{2})^2-(\frac{1}{2})^2}}dx = \int \frac{1}{\sqrt{t^2-(\frac{1}{2})^2}}dt$ .............by using formula $\int \frac{1}{\sqrt{x^2-a^2}}=\log\left | x+\sqrt{x^2+a^2} \right |$
$\\= \log \left | t+\sqrt{t^2-(1/2)^2} \right |+C\\ = \log \left | (x-\frac{3}{2})+\sqrt{x^2-3x+2} \right |+C$

Question:14 Integrate the functions $\frac{1}{\sqrt { 8 + 3 x - x ^ 2 }}$

Answer:

We can write denominator as
$\\=8-(x^2-3x+\frac{9}{4}-\frac{9}{4})\\ =\frac{41}{4}-(x-\frac{3}{2})^2$

therefore
$\Rightarrow \int \frac{1}{\sqrt{8+3x-x^2}}dx= \int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^2}}$
let $x-3/2 = t \Rightarrow dx =dt$

$\therefore$
$\\=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^2-t^2}}dt\\ =\sin^{-1}(\frac{t}{\frac{\sqrt{41}}{2}})+C\\ =\sin^{-1}(\frac{2x-3}{\sqrt{41}})+C$

Question:15 Integrate the functions $\frac{1}{\sqrt {(x-a)( x-b )}}$

Answer:

(x-a)(x-b) can be written as $x^2-(a+b)x+ab$
$\\x^2-(a+b)x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}\\ (x-\frac{(a+b)}{2}^2)^2-\frac{(a-b)^2}{4}$

$\Rightarrow \int\frac{1}{\sqrt{(x-a)(x-b)}}dx=\int \frac{1}{\sqrt{(x-\frac{(a+b)}{2}^2)^2-\frac{(a-b)^2}{4}}}dx$
let
$x-\frac{(a+b)}{2}=t \Rightarrow dx =dt$
So,
$\\=\int \frac{1}{\sqrt{t^2-(\frac{a-b}{2})^2}}dt\\ =\log \left | t+\sqrt{t^2-(\frac{a-b}{2})^2} \right |+C\\ =\log \left | x-(\frac{a+b}{2})+\sqrt{(x-a)(x-b)} \right |+C$

Question:16 Integrate the functions $\frac{4x+1 }{\sqrt {2x ^ 2 + x -3 }}$

Answer:

let
$\\4x+1 = A\frac{d}{dx}(2x^2+x-3)+B\\ 4x+1=A(4x+1)+B\\ 4x+1=4Ax+A+B$

By equating the coefficient of x and constant term on each side, we get
A = 1 and B=0

Let $(2x^2+x-3) = t\Rightarrow (4x+1)dx =dt$

$\therefore \int \frac{4x+1}{\sqrt{2x^2+x-3}}dx= \int\frac{1}{\sqrt{t}}dt$
$\\= 2\sqrt{t}+C\\ =2\sqrt{2x^2+x-3}+C$

Question:17 Integrate the functions $\frac{x+ 2 }{\sqrt { x ^2 -1 }}$

Answer:

let $x+2 =A\frac{d}{dx}(x^2-1)+B=A(2x)+B$
By comparing the coefficients and constant term on both sides, we get;

A=1/2 and B=2
then $x+2 = \frac{1}{2}(2x)+2$

$\int \frac{x+2}{\sqrt{x^2-1}}dx =\int\frac{1/2(2x)+2}{x^2-1}dx$
$\\=\frac{1}{2}\int\frac{(2x)}{\sqrt{x^2-1}}dx+\int \frac{2}{\sqrt{x^2-1}}dx\\ =\frac{1}{2}[2\sqrt{x^2-1}]+2\log\left | x+\sqrt{x^2-1} \right |+C\\ =\sqrt{x^2-1}+2\log\left | x+\sqrt{x^2-1} \right |+C$

Question:18 Integrate the functions $\frac{5x -2 }{1+ 2x +3x^2 }$

Answer:

let
$\\5x+2 = A\frac{d}{dx}(1+2x+3x^2)+B\\ 5x+2= A(2+6x)+B = 2A+B+6Ax$
By comparing the coefficients and constants we get the value of A and B

A = $5/6$ and B = $-11/3$

NOW,
$I = \frac{5}{6}\int \frac{6x+2}{3x^2+2x+1}dx-\frac{11}{3}\int \frac{dx}{3x^2+2x+1}$
$I = I_{1}-\frac{11}{3}I_{2}$ ...........................(i)

put $3x^2+2x+1 =t \Rightarrow (6x+2)dx =dt$
Thus
$I_{1}=\frac{5}{6}\int \frac{dt}{t} =\frac{5}{6}\log t =\frac{5}{6}\log (3x^2+2x+1)+c1$
$I_{2}= \int \frac{dx}{3x^2+2x+1} = \frac{1}{3}\int\frac{dx}{(x+1/3)^2+(\sqrt{2}/3)^2}$
$\\=\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt{2}})+c2$

$\therefore I = I_1+I_2$
$I = \frac{5}{6}\log(3x^2+2x+1)-\frac{11}{3}\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt2})+C$

Question:19 Integrate the functions $\frac{6x + 7 }{\sqrt {( x-5 )( x-4)}}$

Answer:

let
$6x+7 = A\frac{d}{dx}(x^2-9x+20)+B =A(2x-9)+B$
By comparing the coefficients and constants on both sides, we get
A =3 and B =34

$I =\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx = \int \frac{3(2x+9)}{\sqrt{x^2-9x+20}}dx+34\int\frac{dx}{\sqrt{x^2-9x+20}}$ $I = I_1+I_2$ ....................................(i)

Considering $I_1$

$I_1 =\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx$ let $x^2-9x+20 = t \Rightarrow (2x-9)dx =dt$

$I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{x^2-9x+20}$

Now consider $I_2$

$I_2=\int \frac{dx}{\sqrt{x^2-9x+20}}$
here the denominator can be also written as
Dr = $(x-\frac{9}{2})^2-(\frac{1}{2})^2$

$\therefore I_2 = \int \frac{dx}{\sqrt{(x-\frac{9}{2})^2-(\frac{1}{2})^2}}$
$\\= \log\left | (x-\frac{9}{2})^2+\sqrt{x^2-9x+20} \right |$

Now put the values of $I_1$ and $I_2$ in eq (i)

$\\I = 3I_1+34I_2\\ I=6\sqrt{x^2-9x+20}+34\log\left | (x-\frac{9}{2})+\sqrt{x^2-9x+20} \right |+C$

Question:20 Integrate the functions $\frac{x +2 }{\sqrt { 4x - x ^ 2 }}$

Answer:

let
$x+2 = A\frac{d}{dx}(4x-x^2)+B = A(4-2x)+B$
By equating the coefficients and constant term on both sides we get

A = -1/2 and B = 4

(x+2) = -1/2(4-2x)+4

$\\\therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx = -\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^2}}+4\int \frac{dx}{\sqrt{4x-x^2}}\\ \ I =\frac{-1}{2}I_1+4I_2$ ....................(i)

Considering $I_1$
$\int \frac{4-2x}{\sqrt{4x-x^2}}dx$
let $4x-x^2 =t \Rightarrow (4-2x)dx =dt$
$I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{4x-x^2}$
now, $I_2$

$I_2 =\int \frac{dx}{\sqrt{4x-x^2}} = \int \frac{dx}{\sqrt{2^2-(x-2)^2}}$
$=\sin^{-1}(\frac{x-2}{2})$

put the value of $I_1$ and $I_2$

$I =-\sqrt{4x-x^2}+4\sin^{-1}(\frac{x-2}{2})+C$

Question:21 Integrate the functions $\frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}$

Answer:

$\frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}$
$\int \frac{x+2}{\sqrt{x^2+2x+3}}dx = \frac{1}{2}\int \frac{2(x+2)}{\sqrt{x^2+2x+3}}dx$
$\\= \frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\frac{1}{2}\int \frac{2}{\sqrt{x^2+2x+3}}dx\\ =\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\int \frac{1}{\sqrt{x^2+2x+3}}dx\\ I=\frac{1}{2}I_1+I_2$ ...........(i)

take $I_1$

$\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx$
let $x^2+2x+3 = t \Rightarrow (2x+2)dx =dt$

$I_1=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+2x+3}$
considering $I_2$

$= \int \frac{dx}{\sqrt{x^2+2x+3}}= \int \frac{dx}{\sqrt{(x+1)^2+(\sqrt{2})^2}}$
$= \log \left | (x+1)+\sqrt{x^2+2x+3} \right |$
putting the values in equation (i)

$I=\sqrt{x^2+2x+3} +\log \left | (x+1)+\sqrt{x^2+2x+3} \right |+C$

Question:22 Integrate the functions $\frac{x + 3 }{x ^ 2 - 2x - 5 }$

Answer:

Let $(x+3) =A\frac{d}{dx}(x^2-2x+5)+B= A(2x-2)+B$

By comparing the coefficients and constant term, we get;

A = 1/2 and B =4

$\\\int \frac{x+3}{x^2-2x+5}dx = \frac{1}{2}\int \frac{2x-2}{x^2-2x+5}dx +4\int \frac{1}{x^2-2x+5}dx\\ I=I_1+I_2$ ..............(i)

$\\\Rightarrow I_1\\ =\int \frac{2x-2}{x^2-2x-5}dx$
put $x^2-2x-5 =t \Rightarrow (2x-2)dx =dt$

$=\int \frac{dt}{t} = \log t = \log (x^2-2x-5)$

$\\\Rightarrow I_2\\ = \int \frac{1}{x^2-2x-5}dx\\ =\int \frac{1}{(x-1)^2+(\sqrt{6})^2}dx\\ =\frac{1}{2\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})$

$I=I_1+I_2$

$=\frac{1}{2}\log\left | x^2-2x-5 \right |+\frac{2}{\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})+C$

Question:23 Integrate the functions $\frac{5x + 3 }{\sqrt { x^2 + 4x +10 }}$

Answer:

let
$5x+3 = A\frac{d}{dx}(x^2+4x+10)+B = A(2x+4)+B$
On comparing, we get

A =5/2 and B = -7

$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx = \frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx-7\int \frac{dx}{\sqrt{x^2+4x+10}}dx$ $I = 5/2I_1-7I_2$ ...........................................(i)

$\\\Rightarrow I_1\\ \int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$
put
$x^2+4x+10= t \Rightarrow (2x+4)dx = dt$

$=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+4x+10}$

$\\\Rightarrow I_2\\ =\int \frac{1}{\sqrt{x^2+4x+10}}dx \\ =\int \frac{1}{\sqrt{(x+2)^2+(\sqrt{6})^2}}dx\\ =\log \left | (x+2)+\sqrt{x^2+4x+10} \right |$

$I = 5\sqrt{x^2+4x+10}-7\log\left | (x+2)+\sqrt{x^2+4x+10} \right |+C$

Question:24 Choose the correct answer

$\int \frac{dx }{x^2 + 2x +2 }\: \: equals$

$(A) x \tan^{-1} (x + 1) + C\\\\ (B) \tan^{-1} (x + 1) + C\\\\ (C) (x + 1) \tan^{-1}x + C \\\\ (D) \tan^{-1}x + C$

Answer:

The correct option is (B)

$\int \frac{dx }{x^2 + 2x +2 }\: \: equals$
the denominator can be written as $(x+1)^2+1$
now, $\int \frac{dx}{(x+1)^2+1} = tan^{-1}(x+1)+C$

Question:25 Choose the correct answer $\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals$

$A) \frac{1}{9} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\B ) \frac{1}{2} \sin ^{-1}\left ( \frac{8x-9}{9} \right )+ C \\\\ C) \frac{1}{3} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\ D ) \frac{1}{2} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C$

Answer:

The following integration can be done as

$\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals$
$\int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x)}}= \int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x+81/64-81/64)}}dx$
$\\= \int \frac{1}{\sqrt{-4[(x-9/8)^2-(9/8)^2]}}dx\\ =\frac{1}{2}\int \frac{1}{\sqrt{-(x-9/8)^2+(9/8)^2}}dx\\ =\frac{1}{2}[\sin^{-1}(\frac{x-9/8}{9/8})]+C\\ =\frac{1}{2}\sin^{-1}(\frac{8x-9}{9})+C$

The correct option is (B)

NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.5

Question:1 Integrate the rational functions $\frac{x }{( x +1)( x+2)}$

Answer:

Given function $\frac{x }{( x +1)( x+2)}$

Partial function of this function:

$\frac{x }{( x +1)( x+2)} = \frac{A}{(x+1)}+\frac{B}{(x+2)}$

$\implies x = A(x+2)+B(x+1)$

Now, equating the coefficients of x and constant term, we obtain

$A+B =1$

$2A+B =0$

On solving, we get

$A=-1\ and\ B =2$

$\therefore \frac{x}{(x+1)(x+2)} = \frac{-1}{(x+1)}+\frac{2}{(x+2)}$

$\implies \int \frac{x}{(x+1)(x+2)} dx =\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} dx$

$=-\log|x+1| +2\log|x+2| +C$

$=\log(x+2)^2-\log|x+1|+C$

$=\log\frac{(x+2)^2}{(x+1)}+C$

Question:2 Integrate the rational functions $\frac{1}{x^2 -9 }$

Answer:

Given function $\frac{1}{x^2 -9 }$

The partial function of this function:

$\frac{1}{(x+3)(x-3)}= \frac{A}{(x+3)}+\frac{B}{(x-3)}$

$1 = A(x-3)+B(x+3)$

Now, equating the coefficients of x and constant term, we obtain

$A+B =1$

$-3A+3B =1$

On solving, we get

$A=-\frac{1}{6}\ and\ B =\frac{1}{6}$

$\frac{1}{(x+3)(x-3)}= \frac{-1}{6(x+3)} +\frac{1}{6(x-3)}$

$\int \frac{1}{(x^2-9)}dx = \int \left ( \frac{-1}{6(x+3)}+\frac{1}{6(x-3)} \right )dx$

$=-\frac{1}{6}\log|x+3| +\frac{1}{6}\log|x-3| +C$

$= \frac{1}{6}\log\left | \frac{x-3}{x+3} \right |+C$

Question:3 Integrate the rational functions $\frac{3x -1}{( x-1)(x-2)(x-3)}$

Answer:

Given function $\frac{3x -1}{( x-1)(x-2)(x-3)}$

Partial function of this function:

$\frac{3x -1}{( x-1)(x-2)(x-3)}= \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ .(1)

Now, substituting $x=1,2,\ and\ 3$ respectively in equation (1), we get

$A =1,\ B=-5,\ and\ C=4$

$\therefore \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{(x-1)} -\frac{5}{(x-2)}+\frac{4}{(x-3)}$

That implies $\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)} \right \}dx$

$= \log|x-1|-5\log|x-2|+4\log|x-3|+C$

Question:4 Integrate the rational functions $\frac{x }{( x-1)(x-2)(x-3)}$

Answer:

Given function $\frac{x }{( x-1)(x-2)(x-3)}$

Partial function of this function:

$\frac{x }{( x-1)(x-2)(x-3)}= \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$x = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ .....(1)

Now, substituting $x=1,2,\ and\ 3$ respectively in equation (1), we get

$A =\frac{1}{2},\ B=-2,\ and\ C=\frac{3}{2}$

$\therefore \frac{x}{(x-1)(x-2)(x-3)} = \frac{1}{2(x-1)} -\frac{2}{(x-2)}+\frac{3}{2(x-3)}$

That implies $\int \frac{x}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)} \right \}dx$

$= \frac{1}{2}\log|x-1|-2\log|x-2|+\frac{3}{2}\log|x-3|+C$

Question:5 Integrate the rational functions $\frac{2x}{x^2 + 3x +2 }$

Answer:

Given function $\frac{2x}{x^2 + 3x +2 }$

Partial function of this function:

$\frac{2x}{x^2 + 3x +2 }= \frac{A}{(x+1)}+\frac{B}{(x+2)}$

$2x = A(x+2)+B(x+1)$ ...........(1)

Now, substituting $x=-1\ and\ -2$ respectively in equation (1), we get

$A ={-2},\ B=4$

$\frac{2x}{x^2 + 3x +2 }= \frac{-2}{(x+1)}+\frac{4}{(x+2)}$

That implies $\int \frac{2x}{x^2 + 3x +2 }dx= \int \left \{ \frac{-2}{(x+1)}+\frac{4}{(x+2)} \right \}dx$

$=4\log|x+2| -2\log|x+1| +C$

Question:6 Integrate the rational functions $\frac{1- x^2 }{ x ( 1- 2x )}$

Answer:

Given function $\frac{1- x^2 }{ x ( 1- 2x )}$

Integral is not a proper fraction so,

Therefore, on dividing $(1-x^2)$ by $x(1-2x)$ , we get

$\frac{1- x^2 }{ x ( 1- 2x )} = \frac{1}{2} +\frac{1}{2}\left ( \frac{2-x}{x(1-2x)} \right )$

Partial function of this function:

$\frac{2-x}{x(1-2x)} =\frac{A}{x}+\frac{B}{(1-2x)}$

$(2-x) =A(1-2x)+Bx$ ...........(1)

Now, substituting $x=0\ and\ \frac{1}{2}$ respectively in equation (1), we get

$A =2,\ B=3$

$\therefore \frac{2-x}{x(1-2x)} = \frac{2}{x}+\frac{3}{1-2x}$

No, substituting in equation (1) we get

$\frac{1-x^2}{(1-2x)} = \frac{1}{2}+\frac{1}{2}\left \{ \frac{2}{3}+\frac{3}{(1-2x)} \right \}$

$\implies \int \frac{1-x^2}{x(1-2x)}dx =\int \left \{ \frac{1}{2}+\frac{1}{2}\left ( \frac{2}{x}+\frac{3}{1-2x} \right ) \right \}dx$

$=\frac{x}{2}+\log|x| +\frac{3}{2(-2)}\log|1-2x| +C$

$=\frac{x}{2}+\log|x| -\frac{3}{4}\log|1-2x| +C$

Question:7 Integrate the rational functions $\frac{x }{( x^2+1 )( x-1)}$

Answer:

Given function $\frac{x }{( x^2+1 )( x-1)}$

Partial function of this function:

$\frac{x }{( x^2+1 )( x-1)} = \frac{Ax+b}{(x^2+1)} +\frac{C}{(x-1)}$

$x = (Ax+B)(x-1)+C(x^2+1)$

$x=Ax^-Ax+Bc-B+Cx^2+C$

Now, equating the coefficients of $x^2, x$ and the constant term, we get

$A+C = 0$

$-A+B =1$ and $-B+C = 0$

On solving these equations, we get

$A = -\frac{1}{2}, B= \frac{1}{2},\ and\ C=\frac{1}{2}$

From equation (1), we get

$\therefore \frac{x}{(x^2+1)(x-1)} = \frac{\left ( -\frac{1}{2}x+\frac{1}{2} \right )}{x^2+1}+\frac{\frac{1}{2}}{(x-1)}$

$\implies \int \frac{x}{(x^2+1)(x-1)}$

$=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2} \int \frac{1}{x-1}dx$

$=- \frac{1}{4} \int \frac{2x}{x^2+1} dx +\frac{1}{2} \tan^{-1}x + \frac{1}{2} \log|x-1| +C$

Now, consider $\int \frac{2x}{x^2+1} dx$ ,

and we will assume $(x^2+1) = t \Rightarrow 2xdx =dt$

So, $\int \frac{2x}{x^2+1}dx = \int \frac{dt}{t} =\log|t| = \log|x^2+1|$

$\therefore \int \frac{x}{(x^2+1)(x-1)} =-\frac{1}{4}\log|x^2+1| +\frac{1}{2}\tan^{-1}x +\frac{1}{2}\log|x-1| +C$ or

$\frac{1}{2}\log|x-1| - \frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x +C$

Question:8 Integrate the rational functions $\frac{x }{( x+1)^2 ( x+2)}$

Answer:

Given function $\frac{x }{( x+1)^2 ( x+2)}$

Partial function of this function:

$\frac{x }{( x+1)^2 ( x+2)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}$

$x = A(x-1)(x+2)+B(x+2)+C(x-1)^2$

Now, putting $x=1$ in the above equation, we get

$B =\frac{1}{3}$

By equating the coefficients of $x^2$ and constant term, we get

$A+C=0$

$-2A+2B+C = 0$

then after solving, we get

$A= \frac{2}{9}\ and\ C=\frac{-2}{9}$

Therefore,

$\frac{x}{(x-1)^2(x+2)} = \frac{2}{9(x-1)}+\frac{1}{3(x-1)^2}-\frac{2}{9(x+2)}$

$\int \frac{x}{(x-1)^2(x+2)}dx= \frac{2}{9}\int \frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1)^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx$

$= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C$

$\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C$

Question:9 Integrate the rational functions $\frac{3x+ 5 }{x^3 - x^2 - x +1 }$

Answer:

Given function $\frac{3x+ 5 }{x^3 - x^2 - x +1 }$

can be rewritten as $\frac{3x+ 5 }{x^3 - x^2 - x +1 } = \frac{3x+5}{(x-1)^2(x+1)}$

Partial function of this function:

$\frac{3x+5}{(x-1)^2(x+1)}= \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}$

$3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)^2$

$3x+5 = A(x^2-1)+B(x+1)+C(x^2+1-2x)$ ................(1)

Now, putting $x=1$ in the above equation, we get

$B =4$

By equating the coefficients of $x^2$ and $x$ , we get

$A+C=0$

$B-2C =3$

then after solving, we get

$A= -\frac{1}{2}\ and\ C=\frac{1}{2}$

Therefore,

$\frac{3x+5}{(x-1)^2(x+1)}= \frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}$

$\int \frac{3x+5}{(x-1)^2(x+1)}dx= \frac{-1}{2}\int \frac{1}{(x-1)}dx+4\int \frac{1}{(x-1)^2} dx+\frac{1}{2}\int \frac{1}{(x+1)}dx$

$= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C$

$=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C$

Question:10 Integrate the rational functions $\frac{2x -3 }{(x^2 -1 )( 2x+3)}$

Answer:

Given function $\frac{2x -3 }{(x^2 -1 )( 2x+3)}$

can be rewritten as $\frac{2x -3 }{(x^2 -1 )( 2x+3)} = \frac{2x-3}{(x+1)(x-1)(2x+3)}$

The partial function of this function:

$\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{A}{(x+1)} +\frac{B}{(x-1)}+\frac{C}{(2x+3)}$

$\Rightarrow (2x-3) =A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)$ $\Rightarrow (2x-3) =A(2x^2+x-3)+B(2x^2+5x+3)+C(x^2-1)$ $\Rightarrow (2x-3) =(2A+2B+C)x^2+(A+5B)x+(-3A+3B-C)$

Equating the coefficients of $x^2\ and\ x$ , we get

$B=-\frac{1}{10},\ A =\frac{5}{2},\ and\ C= -\frac{24}{5}$

Therefore,

$\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{5}{2(x+1)} -\frac{1}{10(x-1)}-\frac{24}{5(2x+3)}$

$\implies \int \frac{2x-3}{(x^2-1)(2x+3)}dx = \frac{5}{2}\int \frac{1}{(x+1)}dx -\frac{1}{10}\int \frac{1}{x-1}dx -\frac{24}{5}\int \frac{1}{(2x+3)}dx$ $= \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{24}{10}\log|2x+3|$

$= \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{12}{5}\log|2x+3|+C$

$= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C$

$=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C$

$= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C$

$\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C$

Question:11 Integrate the rational functions $\frac{5x}{(x+1)(x^2-4)}$

Answer:

Given function $\frac{5x}{(x+1)(x^2-4)}$

can be rewritten as $\frac{5x}{(x+1)(x^2-4)} = \frac{5x}{(x+1)(x+2)(x-2)}$

The partial function of this function:

$\frac{5x }{(x+1)( x+2)(x-2)} = \frac{A}{(x+1)} +\frac{B}{(x+2)}+\frac{C}{(x-2)}$

$\Rightarrow (5x) =A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)$

Now, substituting the value of $x =-1,-2,\ and\ 2$ respectively in the equation above, we get

$A=\frac{5}{3},\ B =\frac{-5}{2},\ and\ C= \frac{5}{6}$

Therefore,

$\frac{5x }{(x+1)( x+2)(x-2)} = \frac{5}{3(x+1)} -\frac{5}{2(x+2)}+\frac{5}{6(x-2)}$

$\implies \int \frac{5x}{(x+1)(x^2-4)}dx = \frac{5}{3}\int \frac{1}{(x+1)}dx -\frac{5}{2}\int \frac{1}{x+2}dx+\frac{5}{6}\int \frac{1}{(x-2)}dx$ $= \frac{5}{3}\log|x+1| -\frac{5}{2}\log|x+2| +\frac{5}{6}\log|x-2|+C$

Question:12 Integrate the rational functions $\frac{x^3 + x +1}{ x^2-1}$

Answer:

Given function $\frac{x^3 + x +1}{ x^2-1}$

As the given integral is not a proper fraction.

So, we divide $(x^3+x+1)$ by $x^2-1$ , we get

$\frac{x^3 + x +1}{ x^2-1} = x+\frac{2x+1}{x^2-1}$

can be rewritten as $\frac{2x+1}{x^2-1} =\frac{A}{(x+1)} +\frac{B}{(x-1)}$

$2x+1 ={A}{(x-1)} +{B}{(x+1)}$ ....................(1)

Now, substituting $x =1\ and\ x=-1$ in equation (1), we get

$A =\frac{1}{2}\ and\ B=\frac{3}{2}$

Therefore,

$\frac{x^3+x+1 }{(x^2-1)} =x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$

$\implies \int \frac{x^3+x+1 }{(x^2-1)}dx =\int xdx +\frac{1}{2}\int \frac{1}{(x+1)} dx+\frac{3}{2} \int \frac{1}{(x-1)}dx$

$= \frac{x^2}{2}+\frac{1}{2}\log|x+1| +\frac{3}{2}\log|x-1| +C$

Question:13 Integrate the rational functions $\frac{2}{(1-x)(1+ x^2)}$

Answer:

Given function $\frac{2}{(1-x)(1+ x^2)}$

can be rewritten as $\frac{2}{(1-x)(1+ x^2)} = \frac{A}{(1-x)}+\frac{Bx+C}{1+x^2}$

$2 =A(1+x^2)+(Bx+C)(1-x)$ ....................(1)

$2 =A +Ax^2 +Bx-Bx^2+C-Cx$

Now, equating the coefficient of $x^2, x,$ and constant term, we get

$A-B= 0$ , $B-C = 0$ , and $A+C =2$

Solving these equations, we get

$A=1, B=1,\ and\ C=1$

Therefore,

$\therefore \frac{2}{(1-x)(1+ x^2)} = \frac{1}{(1-x)}+\frac{x+1}{1+x^2}$

$\implies \int \frac{2}{(1-x)(1+ x^2)}dx =\int \frac{1}{(1-x)} dx+ \int \frac{x}{1+x^2}dx +\int \frac{1}{1+x^2}dx$ $= -\int \frac{1}{x-1}dx +\frac{1}{2}\int \frac{2x}{1+x^2}dx +\int\frac{1}{1+x^2}dx$

$=-\log|x-1| +\frac{1}{2}\log|1+x^2| +\tan^{-1}x+C$

Question:14 Integrate the rational functions $\frac{3x-1}{(x+2)^2}$

Answer:

Given function $\frac{3x-1}{(x+2)^2}$

can be rewritten as $\frac{3x-1}{(x+2)^2} = \frac{A}{(x+2)}+\frac{B}{(x+2)^2}$

$3x-1 = A(x+2)+B$

Now, equating the coefficient of $x$ and constant term, we get

$A=3$ and $2A+B = -1$ ,

Solving these equations, we get

$B=-7$

Therefore,

$\frac{3x-1}{(x+2)^2} = \frac{3}{(x+2)}-\frac{7}{(x+2)^2}$

$\implies \int\frac{3x-1}{(x+2)^2}dx = 3 \int \frac{1}{(x+2)}dx-7\int \frac{x}{(x+2)^2}dx$

$\implies 3\log|x+2| -7\left ( \frac{-1}{(x+2)}\right )+C$

$\implies 3\log|x+2| + \frac{7}{(x+2)} +C$

Question:15 Integrate the rational functions $\frac{1}{x^4 -1 }$

Answer:

Given function $\frac{1}{x^4 -1 }$

can be rewritten as $\frac{1}{x^4 -1 } = \frac{1}{(x^2-1)(x^2+1)} =\frac{1}{(x+1)(x-1)(1+x^2)}$

The partial fraction of above equation,

$\frac{1}{(x+1)(x-1)(1+x^2)} = \frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{Cx+D}{(x^2+1)}$

$1 = A(x-1)(x^2+1) +B(x+1)(x^2+1)+(Cx+D)(x^2-1)$

$1 = A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+Cx^3+Dx^2-Cx-D$ $1 = (A+B+C)x^3 +(-A+B+D)x^2+(A+B-C)x+(-A+B-D)$

Now, equating the coefficient of $x^3,x^2,x$ and constant term, we get

$A+B+C = 0$ and $-A+B+D = 0$

$A+B-C = 0$ and $-A+B-D = 1$

Solving these equations, we get

$A= -\frac{1}{4}, B=\frac{1}{4},C=0,\ and\ D = -\frac{1}{2}$

Therefore,

$\frac{1}{x^4-1} = \frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x^2+1)}$

$\implies \int \frac{1}{x^4-1}dx = -\frac{1}{4}\log|x-1| +\frac{1}{4}\log|x-1| -\frac{1}{2}\tan^{-1}x +C$

$= \frac{1}{4}\log|\frac{x-1}{x+1}| -\frac{1}{2}\tan^{-1}x +C$

Question:16 Integrate the rational functions $\frac{1}{x ( x^n+1)}$

[Hint: multiply numerator and denominator by $x ^{n-1}$ and put $x ^n = t$ ]

Answer:

Given function $\frac{1}{x ( x^n+1)}$

Applying Hint multiplying numerator and denominator by $x^{n-1}$ and putting $x^n =t$

$\frac{1}{x ( x^n+1)} = \frac{x^{n-1}}{x^{n-1}x(x^n+1)} = \frac{x^{n-1}}{x^n(x^n+1)}$

Putting $x^n =t$

$\therefore x^{n-1}dx =dt$

can be rewritten as $\int \frac{1}{x ( x^n+1)}dx =\int \frac{x^{n-1}}{x^n(x^n+1)}dx = \frac{1}{n} \int \frac{1}{t(t+1)}dt$

Partial fraction of above equation,

$\frac{1}{t(t+1)} =\frac{A}{t}+\frac{B}{(t+1)}$

$1 = A(1+t)+Bt$ ................(1)

Now, substituting $t = 0,-1$ in equation (1), we get

$A=1\ and\ B=-1$

$\therefore \frac{1}{t(t+1)} = \frac{1}{t}- \frac{1}{(1+t)}$

$\implies \int \frac{1}{x(x^n+1)}dx = \frac{1}{n} \int \left \{ \frac{1}{t}-\frac{1}{(t+1)} \right \}dx$

$= \frac{1}{n} \left [ \log|t| -\log|t+1| \right ] +C$

$= -\frac{1}{n} \left [ \log|x^n| -\log|x^n+1| \right ] +C$

$= \frac{1}{n} \log|\frac{x^n}{x^n+1}| +C$

Question:17 Integrate the rational functions $\frac{\cos x }{(1- \sin x )( 2- \sin x )}$

[Hint : Put $\sin x = t$ ]

Answer:

Given function $\frac{\cos x }{(1- \sin x )( 2- \sin x )}$

Applying the given hint: putting $\sin x =t$

We get, $\cos x dx =dt$

$\therefore \int \frac{\cos x }{(1- \sin x )( 2- \sin x )}dx = \int \frac{dt}{(1-t)(2-t)}$

Partial fraction of above equation,

$\frac{1}{(1-t)(2-t)} =\frac{A}{(1-t)}+\frac{B}{(2-t)}$

$1 = A(2-t)+B(1-t)$ ................(1)

Now, substituting $t = 2\ and\ 1$ in equation (1), we get

$A=1\ and\ B=-1$

$\therefore \frac{1}{(1-t)(2-t)} = \frac{1}{(1-t)} - \frac{1}{(2-t)}$

$\implies \int \frac{\cos x }{(1-\sin x)(2-\sin x )}dx = \int \left \{ \frac{1}{1-t}-\frac{1}{(2-t)} \right \}dt$

$= -\log|1-t| +\log|2-t| +C$

$= \log\left | \frac{2-t}{1-t} \right |+C$

Back substituting the value of t in the above equation, we get

$= \log\left | \frac{2-\sin x}{1- \sin x} \right |+C$

Question:18 Integrate the rational functions $\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}$

Answer:

Given function $\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}$

We can rewrite it as: $\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \frac{(4x^2+10)}{(x^2+3)(x^2+4)}$

Partial fraction of above equation,

$\frac{(4x^2+10)}{(x^2+3)(x^2+4)} =\frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+4)}$

$4x^2+10 = (Ax+B)(x^2+4)+(Cx+D)(x^2+3)$

$4x^2+10 = Ax^3+4Ax+Bx^2+4B+Cx^3+3Cx+Dx^2+3D$

$4x^2+10 = (A+C)x^3+(B+D)x^2+(4A+3C)x+(3D+4B)$

Now, equating the coefficients of $x^3, x^2, x$ and constant term, we get

$A+C=0$ , $B+D = 4$ , $4A+3C = 0$ , $4B+3D =10$

After solving these equations, we get

$A= 0, B =-2, C=0,\and\ D=6$

$\therefore \frac{4x^2+10}{(x^2+3)(x^2+4)} = \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)}$

$\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \left ( \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)} \right )$

$\implies \int \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} dx= \int \left \{ 1+ \frac{2}{(x^2+3)} - \frac{6}{(x^2+4)} \right \}dx$

$= \int \left \{ 1+ \frac{2}{(x^2+(\sqrt3)^2)} - \frac{6}{(x^2+2^2)} \right \}dx$

$= x+2\left ( \frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt 3} \right ) - 6\left ( \frac{1}{2}\tan^{-1}\frac{x}{2} \right )+C$

$= x+\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} -3\tan^{-1}\frac{x}{2}+C$

Question:19 Integrate the rational functions $\frac{2x }{( x^2 +1)( x^2 +3)}$

Answer:

Given function $\frac{2x }{( x^2 +1)( x^2 +3)}$

Taking $x^2 = t \Rightarrow 2xdx=dt$

$\therefore \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \frac{dt}{(t+1)(t+3)}$

The partial fraction of above equation,

$\frac{1}{(t+3)(t+3)} = \frac{A}{(t+1)}+\frac{B}{(t+3)}$

$1= A(t+3)+B(t+1)$ ..............(1)

Now, substituting $t = -3\ and\ t = -1$ in equation (1), we get

$A =\frac{1}{2}\ and\ B = -\frac{1}{2}$

$\therefore\frac{1}{(t+3)(t+3)} = \frac{1}{2(t+1)}-\frac{1}{2(t+3)}$

$\implies \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \left \{ \frac{1}{2(t+1)}-\frac{1}{2(t+3)} \right \}dt$

$= \frac{1}{2}\log|t+1|- \frac{1}{2}\log|t+3| +C$

$= \frac{1}{2}\log\left | \frac{t+1}{t+3} \right | +C$

$= \frac{1}{2}\log\left | \frac{x^2+1}{x^2+3} \right | +C$

Question:20 Integrate the rational functions $\frac{1}{x (x^4 -1)}$

Answer:

Given function $\frac{1}{x (x^4 -1)}$

So, we multiply numerator and denominator by $x^3$ , to obtain

$\frac{1}{x (x^4 -1)} = \frac{x^3}{x^4(x^4-1)}$

$\therefore \int \frac{1}{x(x^4-1)}dx =\int\frac{x^3}{x^4(x^4-1)}dx$

Now, putting $x^4 = t$

we get, $4x^3dx =dt$

Taking $x^2 = t \Rightarrow 2xdx=dt$

$\therefore \int \frac{1}{x(x^4-1)}dx =\frac{1}{4}\int \frac{dt}{t(t-1)}$

Partial fraction of above equation,

$\frac{1}{t(t-1)} = \frac{A}{t}+\frac{B}{(t-1)}$

$1= A(t-1)+Bt$ ..............(1)

Now, substituting $t = 0\ and\ t = 1$ in equation (1), we get

$A = -1\ and\ B=1$

$\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}$

$\Rightarrow \int \frac{1}{x(x^4+1)}dx =\frac{1}{4}\int \left \{ \frac{-1}{t}+\frac{1}{t-1} \right \}dt$

$= \frac{1}{4} \left [ -\log|t|+\log|t-1| \right ]+C$

$= \frac{1}{4}\log\left | \frac{t-1}{t} \right |+C$

Back substituting the value of t,

$=\frac{1}{4}\log \left | \frac{x^4-1}{x^4} \right | +C$

Question:21 Integrate the rational functions $\frac{1}{( e ^x-1)}$ [Hint : Put $e ^x= t$ ]

Answer:

Given function $\frac{1}{( e ^x-1)}$

So, applying the hint: Putting $e^x = t$

Then $e^x dx= dt$

$\int \frac{1}{( e ^x-1)}dx = \int\frac{1}{t-1}\times\frac{dt}{t} = \int \frac{1}{t(t-1)}dt$

Partial fraction of above equation,

$\frac{1}{t(t-1)} = \frac{A}{t}+\frac{B}{(t-1)}$

$1= A(t-1)+Bt$ ..............(1)

Now, substituting $t = 0\ and\ t = 1$ in equation (1), we get

$A = -1\ and\ B=1$

$\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}$

$\implies \int \frac{1}{t(t-1)}dt = \log \left | \frac{t-1}{t} \right |+C$

Now, back substituting the value of t,

$= \log \left | \frac{e^x-1}{e^x} \right |+C$

Question:22 Choose the correct answer $\int \frac{x dx }{( x-1)(x-2) } \: \: equals$

$A ) \log |\frac{(x-1)^2}{x-2}| + C \\\\ B) \log |\frac{(x-2)^2}{x-1}| + C \\\\ C ) \log |(\frac{x-1}{x-2})^2| + C \\\\ D ) \log |{(x-1)^2}({x-2})| + C$

Answer:

Given integral $\int \frac{x dx }{( x-1)(x-2) }$

Partial fraction of above equation,

$\frac{x}{(x-1)(x-2)} = \frac{A}{(x-1)}+\frac{B}{(x-2)}$

$x= A(x+2)+B(x-1)$ ..............(1)

Now, substituting $x = 1\ and\ x = 2$ in equation (1), we get

$A = -1\ and\ B=2$

$\therefore \frac{x}{(x-1)(x-2)} = -\frac{1}{(x-1)}+\frac{2}{(x-2)}$

$\implies \int \frac{x}{(x-1)(x-2)}dx = \int \left \{ \frac{-1}{(x-1)}+\frac{2}{(x-2)} \right \}dx$

$= -\log|x-1| +2log|x-2| +C$

$=\log \left | \frac{(x-2)^2}{x-1} \right | +C$

Therefore, the correct answer is B.

Question:23 Choose the correct answer $\int \frac{dx}{x ( x ^2+1)} \: \: equals$

$A ) \log |x| - \frac{1}{2} \log ( x^2 +1 ) + C \\\\ B ) \log |x|+ \frac{1}{2} \log ( x^2 +1 ) + C \\\\ C )- \log |x| + \frac{1}{2} \log ( x^2 +1 ) + C \\\\ D ) \frac{1}{2}\log |x| +\log ( x^2 +1 ) + C$

Answer:

Given integral $\int \frac{dx}{x ( x ^2+1)}$

Partial fraction of above equation,

$\frac{1}{x ( x ^2+1)} = \frac{A}{x}+\frac{Bx+c}{x^2+1}$

$1= A(x^2+1)+(Bx+C)x$

Now, equating the coefficients of $x^2,x,$ and the constant term, we get

$A+B = 0$ , $C=0$ , $A=1$

We have the values, $A = 1\ and\ B=-1,\ and\ C=0$

$\therefore \frac{1}{x ( x ^2+1)} = \frac{1}{x}+\frac{-x}{x^2+1}$

$\implies \int \frac{1}{x ( x ^2+1)}dx =\int \left \{ \frac{1}{x}+\frac{-x}{x^2+1}\right \}dx$

$= \log|x| -\frac{1}{2}\log|x^2+1| +C$

Therefore, the correct answer is A.

NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.6

Question:1 Integrate the functions $x \sin x$

Answer:

Given function is
$f(x)=x \sin x$
We will use integrate by parts method
$\int x\sin x = x.\int \sin xdx - \int(\frac{d(x)}{dx}.\int sin x dx)dx\\ \\ \int x\sin x = x.(-\cos x)- \int (1.(-\cos x))dx\\ \\ \int x\sin x= -x\cos x+\sin x + C$
Therefore, the answer is $-x\cos x+\sin x + C$

Question:2 Integrate the functions $x \sin 3x$

Answer:

Given function is
$f(x)=x \sin 3x$
We will use integration by parts method
$\int x\sin 3x = x.\int \sin 3xdx - \int(\frac{d(x)}{dx}.\int sin 3x dx)dx\\ \\ \int x\sin 3x = x.(\frac{-\cos 3x}{3})- \int (1.(\frac{-\cos 3x}{3}))dx\\ \\ \int x\sin 3x= -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C$

Therefore, the answer is $-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C$

Question:3 Integrate the functions $x ^ 2 e ^x$

Answer:

Given function is
$f(x)=x^2e^x$
We will use integration by parts method
$\int x^2e^x= x^2.\int e^xdx - \int(\frac{d(x^2)}{dx}.\int e^x dx)dx\\ \\ \int x^2e^x = x^2.e^x- \int (2x.e^x)dx\\$
Again use integration by parts in $\int (2x.e^x)dx\\$
$\int (2x.e^x)dx = 2x.\int e^x dx - \int (\frac{d(2x)}{dx}.\int e^xdx)dx\\ \int 2x.e^x dx = 2xe^x- \int 2.e^xdx\\ \int 2x.e^x dx = 2xe^x- 2e^x$
Put this value in our equation
we will get,
$\int x^2.e^x dx =x^2e^x -2xe^x+ 2e^x + C\\ \int x^2.e^x dx = e^x(x^2-2x+2)+ C$

Therefore, answer is $e^x(x^2-2x+2)+ C$

Question:4 Integrate the functions $x \log x$

Answer:

Given function is
$f(x)=x.\log x$
We will use integration by parts method
$\int x.\log xdx= \log x.\int xdx - \int(\frac{d(\log x)}{dx}.\int x dx)dx\\ \\ \int x\log xdx = \log x.\frac{x^2}{2}- \int (\frac{1}{x}.\frac{x^2}{2})dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \frac{x^2}{4}+ C$

Therefore, the answer is $\frac{x^2}{2}\log x- \frac{x^2}{4}+ C$

Question:5 Integrate the functions $x \log 2x$

Answer:

Given function is
$f(x)=x.\log 2 x$
We will use integration by parts method
$\int x.\log 2xdx= \log 2x.\int xdx - \int(\frac{d(\log 2x)}{dx}.\int x dx)dx\\ \\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int (\frac{2}{2x}.\frac{x^2}{2})dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C$

Therefore, the answer is $\log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C$

Question:6 Integrate the functions $x^ 2 \log x$

Answer:

Given function is
$f(x)=x^2.\log x$
We will use integration by parts method
$\int x^2.\log xdx= \log x.\int x^2dx - \int(\frac{d(\log x)}{dx}.\int x^2 dx)dx\\ \\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int (\frac{1}{x}.\frac{x^3}{3})dx\\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int \frac{x^2}{3}dx\\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \frac{x^3}{9}+ C$

Therefore, the answer is $\log x.\frac{x^3}{3}- \frac{x^3}{9}+ C$

Question:7 Integrate the functions $x \sin ^{ -1} x$

Answer:

Given function is
$f(x)=x.\sin^{-1} x$
We will use integration by parts method
$\int x.\sin^{-1} xdx= \sin^{-1} x.\int xdx - \int(\frac{d(\sin^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
Now, we need to integrate $\int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x-\sin^{-1}x \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\sin^{-1}x}{4}+C$
Put this value in our equation

Therefore, the answer is $\int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4}$

Question:8 Integrate the functions $x \tan ^{-1} x$

Answer:

Given function is
$f(x)=x.\tan^{-1} x$
We will use integration by parts method
$\int x.\tan^{-1} xdx= \tan^{-1} x.\int xdx - \int(\frac{d(\tan^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}- \int (\frac{1}{1+x^2}.\frac{x^2}{2})dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( \frac{x^2+1}{1+x^2}-\frac{1}{1+x^2} \right )dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( 1-\frac{1}{1+x^2} \right )dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\left ( x- \tan^{-1}x \right )+C\\ \\ \int x\tan^{-1}xdx = \frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$

Put this value in our equation
$\int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \frac{x}{4\sqrt{1-x^2}}-\frac{\sin^{-1}x}{4}+C\\ \int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x}{4\sqrt{1-x^2}}$

Therefore, the answer is $\frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$

Question:9 Integrate the functions $x\cos ^{ -1} x$

Answer:

Given function is
$f(x)=x.\cos^{-1} x$
We will use integration by parts method
$\int x.\cos^{-1} xdx= \cos^{-1} x.\int xdx - \int(\frac{d(\cos^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}- \int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
Now, we need to integrate $\int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\cos^{-1}x+\cos^{-1}x \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C$
Put this value in our equation
$\int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}-\left ( \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C \right )\\ \\ \int x\cos^{-1} xdx =\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}$

Therefore, the answer is $\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}$

Question:10 Integrate the functions $( \sin ^{-1}x ) ^ 2$

Answer:

Given function is
$f(x)=( \sin ^{-1}x ) ^ 2$
we will use integration by parts method
$\int (\sin^{-1}x)^2= (\sin^{-1}x)^2.\int 1dx-\int \left ( \frac{d( (\sin^{-1}x)^2)}{dx} .\int 1dx\right )dx\\ \\ \int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x-\int \left ( \sin^{-1}.\frac{2x}{\sqrt{1-x^2}} \right )dx\\ \int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \left ( \frac{d(\sin^{-1}x)}{dx}. \int \frac{-2x}{\sqrt{1-x^2}}dx\right ) \right ]\\ \\ . \ \ \ \ \ = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.2\sqrt{1-x^2}- \int \frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]\\ \\ . \ \ \ \ \ = (\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C$
Therefore, answer is $(\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C$

Question:11 Integrate the functions $\frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}$

Answer:

Consider $\int \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}dx =I$

So, we have then: $I = \frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}. \cos^{-1}x dx$

After taking $\cos ^{-1}x$ as a first function and $\left ( \frac{-2x}{\sqrt{1-x^2}} \right )$ as second function and integrating by parts, we get

$I =-\frac{1}{2}\left [ \cos^{-1}x\int\frac{-2x}{\sqrt{1-x^2}}dx - \int\left \{ \left ( \frac{d}{dx}\cos^{-1}x \right )\int \frac{-2x}{\sqrt{1-x^2}}dx \right \}dx \right ]$ $=-\frac{1}{2}\left [ \cos^{-1}x.2{\sqrt{1-x^2}} + \int \frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]$

$=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-\int2dx \right ]$

$=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-2x \right ]+C$

Or, $-\left \sqrt{1-x^2}\cos^{-1}x +x\right +C$

Question:12 Integrate the functions $x \sec ^2 x$

Answer:

Consider $x \sec ^2 x$

So, we have then: $I =\int x\sec^2 x dx$

After taking $x$ as a first function and $\sec^2x$ as second function and integrating by parts, we get

$I =x\int \sec^2 x dx -\int \left \{ \left ( \frac{d}{dx}x \right )\int \sec^2 x dx \right \}dx$

$= x\tan x -\int1.\tan x dx$

$= x\tan x +\log|\cos x | +C$

Question:13 Integrate the functions $\tan ^{-1} x$

Answer:

Consider $\tan ^{-1} x$

So, we have then: $I =\int 1.\tan^{-1}x dx$

After taking $\tan^{-1}x$ as a first function and $1$ as second function and integrating by parts, we get

$I = \tan^{-1}x \int 1dx -\int \left \{ \left ( \frac{d}{dx}\tan^{-1}x \right )\int1.dx \right \}dx$

$= \tan^{-1}x.x -\int \frac{1}{1+x^2}.xdx$

$= x\tan^{-1}x -\frac{1}{2}\int \frac{2x}{1+x^2}dx$

$= x\tan^{-1}x -\frac{1}{2}\log|1+x^2|+C$

$= x\tan^{-1}x -\frac{1}{2}\log(1+x^2)+C$

Question:14 Integrate the functions $x ( \log x )^ 2$

Answer:

Consider $x ( \log x )^ 2$

So, we have then: $I = \int x(\log x)^2 dx$

After taking $(\log x )^2$ as a first function and $x$ as second function and integrating by parts, we get

$I = (\log x )^2 \int xdx -\int \left \{ \left ( \frac{d}{dx} (\log x)^2 \right )\int x.dx \right \}dx$

$= (\log x)^2 .\frac{x^2}{2} - \int \frac{2\log x }{x}.\frac{x^2}{2} dx$

$= (\log x)^2 .\frac{x^2}{2} - \int x\log x dx$

$= (\log x)^2 .\frac{x^2}{2} - \left ( \frac{x^2 \log x }{2} -\frac{x^2}{4} \right )+C$

Question:15 Integrate the functions $( x^2 + 1 ) \log x$

Answer:

Consider $( x^2 + 1 ) \log x$

So, we have then: $I = \int (x^2+1) \log x dx = \int x^2 \log x dx +\int \log x dx$

Let us take $I = I_{1} +I_{2}$ ....................(1)

Where, $I_{1} = \int x^2\log x dx$ and $I_{2} = \int \log x dx$

So, $I_{1} = \int x^2\log x dx$

After taking $\log x$ as a first function and $x^2$ as second function and integrating by parts, we get

$I = \log x \int x^2dx -\int \left \{ \left ( \frac{d}{dx} \log x \right )\int x^2.dx \right \}dx$

$= \log x .\frac{x^3}{3} - \int \frac{1}{x}.\frac{x^3}{3} dx$

$= \log x .\frac{x^3}{3} - \frac{x^3}{9} +C_{1}$ ....................(2)

$I_{2} = \int \log x dx$

After taking $\log x$ as a first function and $1$ as second function and integrating by parts, we get

$I_{2} = \log x \int 1.dx - \int \left \{ \left ( \frac{d}{dx}\log x \right ) \int 1.dx \right \}dx$

$= \log x .x -\int \frac{1}{x}. xdx$

$= x\log x -\int 1 dx$

$= x\log x -x +C_{2}$ ................(3)

Now, using the two equations (2) and (3) in (1) we get,

$I = \frac{x^3}{3}\log x -\frac{x^3}{9} +C_{1} +x\log x - x +C_{2}$

$= \frac{x^3}{3}\log x -\frac{x^3}{9} +x\log x - x +(C_{1}+C_{2})$

$=\left ( \frac{x^3}{3}+x \right ) \log x -\frac{x^3}{9} -x+C$

Question:16 Integrate the functions $e ^ x ( \sin x + \cos x )$

Answer:

Let suppose
$I =$ $e ^ x ( \sin x + \cos x )$
$f(x) = \sin x \Rightarrow f'(x) = \cos x$
we know that,
$I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C$
Thus, the solution of the given integral is given by

$\therefore I = e^x\sin x +C$

Question:17 Integrate the functions $\frac{x e ^x }{( 1+ x )^2}$

Answer:

$\frac{x e ^x }{( 1+ x )^2}$
Let suppose
$I = \int \frac{e^x(x)}{(1+x)^2}dx$
by rearranging the equation, we get
$\Rightarrow \int e^x[\frac{1}{1+x}-\frac{1}{(1+x)^2}]dx$
let
$f(x)=\frac{1}{1+x} \Rightarrow f'(x)= -\frac{1}{(1+x)^2}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$
therefore the solution of the given integral is

$I = \frac{e^x}{1+x}+C$

Question:18 Integrate the functions $e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$

Answer:

Let
$I =e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$
substitute $1 =\sin ^2\frac{x}{2}+\cos^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$

$\\\Rightarrow e^x(\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}})\\ =e^x(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2})\\$
let
$f(x) =\tan\frac{x}{2} \Rightarrow f'(x)=\frac{1}{2}\sec^2\frac{x}{2}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$
Therefore the solution of the given integral is

$I = e^x\tan\frac{x}{2} +C$

Question:19 Integrate the functions $e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$

Answer:

$e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$
It is known that
$\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$

let
$f(x)=\frac{1}{x}\Rightarrow f'(x)=-\frac{1}{x^2}$
Therefore the required solution of the given above integral is
$I = e^x.\frac{1}{x}+C$

Question:20 Integrate the functions $\frac{( x-3)e ^x }{( x-1)^3}$

Answer:

$\frac{( x-3)e ^x }{( x-1)^3}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$

So, By adjusting the given equation, we get
$\int\frac{( x-3)e ^x }{( x-1)^3} =\int e^x(\frac{x-1-2}{(x-1)^3}) =\int e^x({\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3})}dx$

to let
$f(x)=\frac{1}{(x-1)^2}\Rightarrow f'(x)=-\frac{2}{(x-1)^3}$
Therefore the required solution of the given $I=\frac{e^x}{(x-1)^2}+C$ integral is

Question:21 Integrate the functions $e ^{ 2x } \sin x$

Answer:

Let
$I =e ^{ 2x } \sin x$
By using integrating by parts, we get

$\\=\sin x\int e ^{ 2x }dx-\int(\frac{d}{dx}\sin x.\int e^{2x}dx)\ dx\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}\int e^{2x}.\cos x\ dx\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x\int e^{2x}dx-\int (\frac{d}{dx}\cos x.\int e^{2x}dx)\ dx]\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x.\frac{e^x}{2}+\frac{1}{2}\int e^{2x}\sin x dx]\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}-\frac{1}{4}I\\ \Rightarrow \frac{5}{4}I =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}\\ I = \frac{e^{2x}}{5}[2\sin x-\cos x]+C$

Question:22 Integrate the functions $\sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )$

Answer:

$\sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )$

$\int \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
let $x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta$

$\\=\int\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int2\theta \sec^2\theta d\theta\\$
Taking $\theta$ as a first function and $\sec^2\theta$ as a second function, by using by parts method

$\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]+C\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]+C\\ =2x\tan^{-1}x+2\log (1+x^2)^{-1/2}\\ =2x\tan^{-1}x-\log(1+x^2)+C$

Question:23 Choose the correct answer

$\int x ^ 2 e ^{x ^3 } dx \: \: equals$

$A ) \frac{1}{3} e ^{x^3} + C \\\\ B) \frac{1}{3} e ^{x^2} + C \\\\ C ) \frac{1}{2} e ^{x^3} + C \\\\ D ) \frac{1}{2} e ^{x^2} + C$

Answer:

the integration can be done ass follows

let $x^3 =t\Rightarrow 3x^2dx=dt$
$\Rightarrow I =\frac{1}{3}\int e^tdt =\frac{1}{3}e^t+C=\frac{1}{3}e^x^3+C$

Question:24 Choose the correct answer

$\int e ^ x \sec ( 1+ \tan x ) dx \: \: \: equals$

$A ) e ^ x \cos x + C \\\\ B) e ^ x \sec x + C \\\\ C ) e ^ x \sin x + C\\\\D ) e ^ x \tan x + C$

Answer:

we know that,
$I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C$
from above integral
let
$f(x)=\sec x\Rightarrow f'(x)= \sec x.\tan x$
thus, the solution of the above integral is
$I=e^x\sec x+C$

NCERT class 12 maths ch 7 question answer Exercise: 7.7

Question:1 Integrate the functions in Exercises 1 to 9.

$\sqrt{4 - x^2}$

Answer:

Given function $\sqrt{4 - x^2}$ ,

So, let us consider the function to be;

$I = \int \sqrt{4-x^2}dx$

$= \int \sqrt{(2)^2-x^2}dx$

Then it is known that, $= \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

Therefore, $I = \frac{x}{2}\sqrt{4-x^2} +\frac{4}{2}\sin^{-1}{\frac{x}{2}}+C$

$= \frac{x}{2}\sqrt{4-x^2} +2\sin^{-1}{\frac{x}{2}}+C$

Question:2 Integrate the functions in Exercises 1 to 9.

$\sqrt{1 - 4x^2}$

Answer:

Given function to integrate $\sqrt{1 - 4x^2}$

Now we can rewrite as

$= \int \sqrt{1 - (2x)^2}dx$

As we know the integration of this form is $\left [ \because \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a} \right ]$

$= \frac{(\frac{2x}{2})\sqrt{1^2-(2x)^2}+\frac{1^2}{2}\sin^{-1}\frac{2x}{1}}{2\rightarrow Coefficient\ of\ x\ in\ 2x} +C$

$= \frac{1}{2}\left [ x\sqrt{1-4x^2}+\frac{1}{2}\sin^{-1}2x \right ]+C$

$= \frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}2x+C$

Question:3 Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 4x + 6}$

Answer:

Given function $\sqrt{x^2 + 4x + 6}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2 + 4x + 6}dx$

$= \int\sqrt{(x^2 + 4x + 4)+2}dx = \int\sqrt{(x + 2)^2 +(\sqrt2)^2}dx$

And we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

$\Rightarrow I = \frac{x+2}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C$

$= \frac{x+2}{2}\sqrt{x^2+4x+6}+\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C$

Question:4 Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 4x +1}$

Answer:

Given function $\sqrt{x^2 + 4x +1}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2 + 4x + 1}dx$

$= \int\sqrt{(x^2 + 4x + 4)-3}dx = \int\sqrt{(x + 2)^2 -(\sqrt3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$\therefore I = \frac{x+2}{2}\sqrt{x^2+4x+1}-\frac{3}{2}\log\left | (x+2)+\sqrt{x^2+4x+1} \right |+C$

Question:5 Integrate the functions in Exercises 1 to 9.

$\sqrt{1-4x-x^2}$

Answer:

Given function $\sqrt{1-4x-x^2}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1-4x-x^2}dx$

$= \int\sqrt{1-(x^2+4x+4-4)}dx = \int\sqrt{1+4 -(x+2)^2}dx$

$= \int\sqrt{(\sqrt5)^2 -(x+2)^2}dx$

And we know that, $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\therefore I = \frac{x+2}{2}\sqrt{1-4x-x^2}+\frac{5}{2}\sin^{-1}\left ( \frac{x+2}{\sqrt5} \right )+C$

Question:6 Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 4x - 5}$

Answer:

Given function $\sqrt{x^2 + 4x - 5}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2+4x-5}dx$

a $= \int\sqrt{(x^2+4x+4)-9}dx = \int\sqrt{(x+2)^2 -(3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}|+C$

$\therefore I = \frac{x+2}{2}\sqrt{x^2+4x-5}-\frac{9}{2}\log\left | (x+2)+ \sqrt{x^2+4x-5} \right |+C$

Question:7 Integrate the functions in Exercises 1 to 9.

$\sqrt{1 + 3x - x^2}$

Answer:

Given function $\sqrt{1 + 3x - x^2}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1+3x-x^2}dx$

$= \int\sqrt{(1-\left ( x^2-3x+\frac{9}{4}-\frac{9}{4} \right )}dx = \int \sqrt{\left ( 1+\frac{9}{4} \right )-\left ( x-\frac{3}{2} \right )^2}dx$ $= \int \sqrt{\left ( \frac{\sqrt{13}}{2} \right )^2-\left ( x-\frac{3}{2} \right )^2}dx$

And we know that, $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\therefore I = \frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}} \right )+C$

$= \frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{2x-3}{\sqrt{13}} \right )+C$

Question:8 Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 3x}$

Answer:

Given function $\sqrt{x^2 + 3x}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2+3x}dx$

$= \int\sqrt{x^2+3x+\frac{9}{4}-\frac{9}{4}}dx$

$= \int\sqrt{\left ( x+\frac{3}{2} \right )^2-\left ( \frac{3}{2} \right )^2 }dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$\therefore I = \frac{x+\frac{3}{2}}{2}\sqrt{x^2+3x}-\frac{\frac{9}{4}}{2}\log \left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C$

$= \frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\log\left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C$

Question:9 Integrate the functions in Exercises 1 to 9.

$\sqrt{1 + \frac{x^2}{9}}$

Answer:

Given function $\sqrt{1 + \frac{x^2}{9}}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1+\frac{x^2}{9}}dx = \frac{1}{3}\int \sqrt{9+x^2}dx$

$= \frac{1}{3}\int \sqrt{3^2+x^2}dx$

And we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

$\therefore I = \frac{1}{3}\left [ \frac{x}{2}\sqrt{x^2+9} +\frac{9}{2}\log|x+\sqrt{x^2+9}| \right ]+C$

$= \frac{x}{6}\sqrt{x^2+9} +\frac{3}{2}\log\left | x+\sqrt{x^2+9} \right |+C$

Question:10 Choose the correct answer in Exercises 10 to 11.

$\int \sqrt{1+x^2}dx$ is equal to

(A) $\frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log\left |\left(x + \sqrt{1+x^2} \right )\right| +C$

(B) $\frac{2}{3}(1+x^2)^{\frac{3}{2}} + C$

(D) $\frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log\left |x + \sqrt{1+x^2} \right| +C$

Answer:

As we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

So, $\int \sqrt{1+x^2}dx = \frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log|x+\sqrt{x^2+1}| +C$

Therefore the correct answer is A.

Question:11 Choose the correct answer in Exercises 10 to 11.

$\int \sqrt{x^2 - 8x+7}dx$ is equal to

(A) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} + 9\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C$

(B) $\frac{1}{2}(x+4)\sqrt{x^2-8x+7} + 9\log\left|x+4+\sqrt{x^2 -8x+7}\right| +C$

(D) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} -\frac{9}{2}\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C$

Answer:

Given integral $\int \sqrt{x^2 - 8x+7}dx$

So, let us consider the function to be;

$I = \int\sqrt{x^2-8x+7}dx =\int\sqrt{(x^2-8x+16)-(9)}dx$

$=\int\sqrt{(x-4)^2-(3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$I = \frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2}\log|(x-4)+\sqrt{x^2-8x+7}| +C$

Therefore the correct answer is D.

NCERT class 12 maths ch 7 question answer - Exercise:7.8

Question:1 Evaluate the following definite integrals as a limit of sums.

$\int_a^b x dx$

Answer:

We know that,
$\int_{a}^{b}f(x)dx = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+...+f(a+(n-1)h)]$
$\therefore \int_{a}^{b}xdx = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[a+(a+h)...(a+2h)..a+(n-1)h]$
$\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[(a+a...a+a)_{n}+(h+2h+3h....(n-1)h)]\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[na+h(1+2+3..+n-1)]\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[na+h(\frac{n(n-1)}{2})]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{n-1}{2}h]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{(n-1)(b-a)}{2n}]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{(1-\frac{1}{n})(b-a)}{2}]\\ = (b-a)[a+\frac{(b-a)}{2}]\\ =(b-a)(b+a)/2\\ =\frac{(b^2-a^2)}{2}$

This is how the integral is evaluated using limit of a sum

Question:2 Evaluate the following definite integrals as limit of sums.

$\int_0^5 (x + 1)dx$

Answer:

We know that
let $I =\int_{0}^{5}(x+1)dx$
$\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}$
Here a = 0, b = 5 and $f(x)=(x+1)$
therefore $h=\frac{5}{n}$

$\int_{0}^{5}(x+1)dx=5\lim_{x\rightarrow \infty }\frac{1}{n}[f(0)+f(5/n)+.....+f((n-1)5/n)]$

$=5\lim_{x\rightarrow \infty }\frac{1}{n}[1+(5/n+1)+....+(1+\frac{5(n-1)}{n})]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[(1+1..+1)_{n}+\frac{5}{n}(1+2+3+...+n-1)]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{5}{n}\frac{n(n-1)}{2}]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{5(n-1)}{2}]\\ =5\lim_{x\rightarrow \infty }[1+\frac{5(1-\frac{1}{n})}{2}]\\ =5[1+\frac{5}{2}]\\ =\frac{35}{2}$

Question:3 Evaluate the following definite integrals as limit of sums.

$\int_2^3 x^2dx$

Answer:

We know that

$\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}$
here a = 2 and b = 3 , so h = 1/n

$\int_{2}^{3}x^2dx=(3-2)\lim_{x\rightarrow \infty }\frac{1}{n}[f(2)+f(2+\frac{1}{n})+f(2+\frac{2}{n})+....+f(2+\frac{(n-1)}{n})]$

$\\=(1)\lim_{x\rightarrow \infty }\frac{1}{n}[2^2+(2+\frac{1}{n})^2+......+(2+\frac{(n-1)}{n})^2]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[(2^2+2^2+...2^2)_{n}+(\frac{1}{n})^2+(\frac{2}{n})^2+....(\frac{n-1}{n})^2+4(\frac{1}{n}+\frac{2}{n}+.....+\frac{n-1}{n})\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[4n+\frac{n(n-1)(2n-1)}{6n^2}+\frac{4}{n}.\frac{n(n-1)}{2}]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[4n+(1-\frac{(1-\frac{1}{n})(2n-1)}{6})+\frac{4(n-1)}{2}]$
$\\=\lim_{x\rightarrow \infty }\frac{1}{n}[4n+(1-\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6})+\frac{4(n-1)}{2}]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}.n[4+(1-\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6})+2-\frac{2}{n}]\\ =4+\frac{2}{6}+2 =\frac{19}{3}$

Question:4 Evaluate the following definite integrals as limit of sums.

$\int_{1}^4(x^2-x)dx$

Answer:

Let
$\\I = \int_{1}^{4}(x^2-x)dx =\int_{1}^{4}x^2dx-\int_{1}^{4}xdx\\ I = I_1-I_2$

$\int_{1}^{4}x^2dx=(4-1)\lim_{x\rightarrow \infty }\frac{1}{n}[f(1)+f(1+h)+f(1+2h)+.....+f(1+(n-1)h)]$

$=(4-1)\lim_{x\rightarrow \infty }\frac{1}{n}[f(1)+f(1+h)+f(1+2h)+.....+f(1+(n-1)h)]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[1^2+(1+\frac{3}{n})^2+(1+2.\frac{3}{n})^2+......+(1+(n-1).\frac{3}{n})^2]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[(1^2+..1^2)_{n}+(\frac{3}{n})^2(1^2+2^2+3^2+....+(n-1)^2)+2.\frac{3}{n}(1+2+3..+n-1)]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{6}{n}(\frac{n(n-1)}{2})]$

$=3\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{6}{n}(\frac{n(n-1)}{2})]\\ =3\lim_{x\rightarrow \infty }[1+\frac{9}{6}(1-\frac{1}{n})(2-\frac{1}{n})+3(1-\frac{1}{n})]\\ =3[1+\frac{9}{6}.2+3]\\ = 21$

for the second part, we already know the general solution of $\int_{a}^{b}xdx = \frac{(b^2-a^2)}{2}$
So, here a = 1 and b = 4
therefore $\int_{1}^{4}xdx = \frac{(4^2-1^2)}{2}=\frac{15}{2}$

So, $I = 21-\frac{15}{2} = \frac{27}{2}$

Question:5 Evaluate the following definite integrals as limit of sums.

. $\int_{-1}^1 e^xdx$

Answer:

let $I = \int_{-1}^{1}e^xdx$
We know that
$\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}$
Here a =-1, b = 1 and $f(x) = e^x$
therefore h = 2/n
$I = 2.\lim_{x\rightarrow \infty }\frac{1}{n}[f(-1)+f(-1+\frac{2}{n})+.....+f(-1+(n-1).\frac{2}{n})]$
$\\ =2.\lim_{x\rightarrow \infty }\frac{1}{n}[e^{-1}+e^{-1+\frac{2}{n}}+e^{-1+2.\frac{2}{n}}+...+e^{-1+(n-1).\frac{2}{n}}]\\ = 2.\lim_{x\rightarrow \infty }\frac{1}{n}[e^{-1}(1+e^{2/n}+e^{4/n}+...+e^{(n-1).\frac{2}{n}})]\\ =$
By using sum of n terms of GP $S =\frac{a(r^n-1)}{r-1}$ ....where a = 1st term and r = ratio

$\\=2\lim_{n\rightarrow \infty }\frac{e^{-1}}{n}[\frac{1.(e^{\frac{2}{n}.n}-1)}{e^\frac{2}{n}-1}]\\ =2\lim_{n\rightarrow \infty }\frac{e^{-1}}{n}(\frac{e^2-1}{e^{2/n}-1})\\ =\frac{e^{-1}(e^2-1)}{\lim_{\frac{2}{n}\rightarrow \infty }\frac{e^{2/n}-1}{2/n}}\\ =\frac{e^2-1}{e}$ .........using $[\lim_{x\rightarrow \infty }(\frac{e^x-1}{x})=1]$

Question:6 Evaluate the following definite integrals as limit of sums.

$\int_0^4(x + e^{2x})dx$

Answer:

It is known that,
$\int_{0}^{4}(x+e^{2x})dx = 4\lim_{x\rightarrow \infty }\frac{1}{n}[f(0)+f(h)+f(2h)+....+f(n-1)h]$
$\\=4\lim_{x\rightarrow \infty }\frac{1}{n}[(0+e^0)+(h+e^2h)+(2h+e^4h)+......+((n-1)h+e^{2(n-1)h})]\\ = 4\lim_{x\rightarrow \infty }\frac{1}{n}[h(1+2+3+.....+n-1)+(\frac{e^{2nh}-1}{e^{2h}-1})]\\ = 4\lim_{x\rightarrow \infty }\frac{1}{n}[\frac{4}{n}(\frac{n(n-1)}{2})+(\frac{e^8-1}{e^{8/n}-1})]$
$\\=4\lim_{x\rightarrow \infty }[4.\frac{1-\frac{1}{n}}{2}+\frac{\frac{e^8-1}{8}}{\frac{e^{8/n}-1}{\frac{8}{n}}}]\\ =4(2)+4[(\frac{e^8-1}{8})]\\ ==8+e^8/2-1/2\\ =\frac{15+e^8}{2}$ ..........................( $\lim_{x\rightarrow 0}\frac{e^x-1}{x}=1$ )

NCERT class 12 maths ch 7 question answer - Exercise:7.9

Question:1 Evaluate the definite integrals in Exercises 1 to 20.

$\int_{-1}^{1} (x+1)dx$

Answer:

Given integral: $I = \int_{-1}^{1} (x+1)dx$

Consider the integral $\int (x+1)dx$

$\int (x+1)dx = \frac{x^2}{2}+x$

So, we have the function of $x$ , $f(x) = \frac{x^2}{2}+x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1)-f(-1)$

$= \left ( \frac{1}{2}+1\right ) - \left (\frac{1}{2}-1 \right ) = \frac{1}{2}+1-\frac{1}{2}+1 = 2$

Question:2 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3\frac{1}{x}dx$

Answer:

Given integral: $I = \int_2^3\frac{1}{x}dx$

Consider the integral $\int_2^3\frac{1}{x}dx$

$\int \frac{1}{x}dx = \log|x|$

So, we have the function of $x$ , $f(x) = \log|x|$

Now, by Second fundamental theorem of calculus, we have

$I = f(3)-f(2)$

$=\log|3|-\log|2| = \log \frac{3}{2}$

Question:3 Evaluate the definite integrals in Exercises 1 to 20.

$\int_1^2(4x^3-5x^2 + 6x +9)dx$

Answer:

Given integral: $I = \int_1^2(4x^3-5x^2 + 6x +9)dx$

Consider the integral $I = \int (4x^3-5x^2 + 6x +9)dx$

$\int (4x^3-5x^2 + 6x +9)dx = 4\frac{x^4}{4} -5\frac{x^3}{3}+6\frac{x^2}{2}+9x$

$= x^4 -\frac{5x^3}{3}+3x^2+9x$

So, we have the function of $x$ , $f(x) = x^4 -\frac{5x^3}{3}+3x^2+9x$

Now, by Second fundamental theorem of calculus, we have

$I = f(2)-f(1)$

$=\left \{ 2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right \} - \left \{ 1^4-\frac{5(1)^3}{3}+3(1)^2+9(1) \right \}$

$=\left \{ 16-\frac{40}{3}+12+18\right \} - \left \{ 1-\frac{5}{3}+3+9 \right \}$

$=\left \{ 46-\frac{40}{3}\right \} - \left \{ 13-\frac{5}{3}\right \}$

$=\left \{ 33-\frac{35}{3} \right \} = \left \{ \frac{99-35}{3} \right \}$

$= \frac{64}{3}$

Question:4 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{4}\sin 2x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{4}\sin 2x dx$

Consider the integral $\int \sin 2x dx$

$\int \sin 2x dx = \frac{-\cos 2x }{2}$

So, we have the function of $x$ , $f(x) = \frac{-\cos 2x }{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4})-f(0)$

$= \frac{-\cos 2(\frac{\pi}{4})}{2} + \frac{\cos 0}{2}$

$=\frac{1}{2} - 0$

$= \frac{1}{2}$

Question:5 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{2}\cos 2x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{2}\cos 2x dx$

Consider the integral $\int \cos 2x dx$

$\int \cos 2x dx = \frac{\sin 2x }{2}$

So, we have the function of $x$ , $f(x) = \frac{\sin 2x }{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{2})-f(0)$

$= \frac{1}{2}\left \{ \sin 2(\frac{\pi}{2}) - \sin 0 \right \}$

$= \frac{1}{2}\left \{ 0 - 0 \right \} = 0$

Question:6 Evaluate the definite integrals in Exercises 1 to 20.

$\int_4^5 e^x dx$

Answer:

Given integral: $\int_4^5 e^x dx$

Consider the integral $\int e^x dx$

$\int e^x dx = e^x$

So, we have the function of $x$ , $f(x) = e^x$

Now, by Second fundamental theorem of calculus, we have

$I = f(5) -f(4)$

$= e^5 -e^4$

$= e^4(e-1)$

Question:7 Evaluate the definite integrals in Exercises 1 to 20.

$\int^\frac{\pi}{4}_0 \tan x dx$

Answer:

Given integral: $\int^\frac{\pi}{4}_0 \tan x dx$

Consider the integral $\int \tan x dx$

$\int \tan x dx = -\log|\cos x |$

So, we have the function of $x$ , $f(x) = -\log|\cos x |$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(0)$

$= -\log\left | \cos \frac{\pi}{4} \right | +\log|\cos 0|$

$= -\log\left | \cos \frac{1}{\sqrt2} \right | +\log|1|$

$= -\log\left | \frac{1}{\sqrt2} \right | + 0 = -\log (2)^{-\frac{1}{2}}$

$= \frac{1}{2}\log (2)$

Question:8 Evaluate the definite integrals in Exercises 1 to 20.

$\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx$

Answer:

Given integral: $\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx$

Consider the integral $\int\textup{cosec}xdx$

$\int\textup{cosec}xdx = \log|cosec x -\cot x |$

So, we have the function of $x$ , $f(x) =\log|cosec x -\cot x |$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(\frac{\pi}{6})$

$= \log|cosec \frac{\pi}{4} -\cot \frac{\pi}{4} | - \log|cosec \frac{\pi}{6} -\cot \frac{\pi}{6} |$

$= \log|\sqrt2 -1 | - \log|2 -\sqrt3 |$

$= \log \left ( \frac{\sqrt2 -1}{2-\sqrt3} \right )$

Question:9 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{dx}{\sqrt{1-x^2}}$

Answer:

Given integral: $\int_0^1\frac{dx}{\sqrt{1-x^2}}$

Consider the integral $\int \frac{dx}{\sqrt{1-x^2}}$

$\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x$

So, we have the function of $x$ , $f(x) = \sin^{-1}x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \sin^{-1}(1) -\sin^{-1}(0)$

$= \frac{\pi}{2} - 0$

$= \frac{\pi}{2}$

Question:10 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{dx}{1 + x^2}$

Answer:

Given integral: $\int_0^1\frac{dx}{1 + x^2}$

Consider the integral $\int\frac{dx}{1 + x^2}$

$\int\frac{dx}{1 + x^2} = \tan^{-1}x$

So, we have the function of $x$ , $f(x) =\tan^{-1}x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \tan^{-1}(1) -\tan^{-1}(0)$

$= \frac{\pi}{4} - 0$

$= \frac{\pi}{4}$

Question:11 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3 \frac{dx}{x^2 -1 }$

Answer:

Given integral: $\int_2^3 \frac{dx}{x^2 -1 }$

Consider the integral $\int \frac{dx}{x^2 -1 }$

$\int \frac{dx}{x^2 -1 } = \frac{1}{2}\log\left | \frac{x-1}{x+1} \right |$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\log\left | \frac{x-1}{x+1} \right |$

Now, by Second fundamental theorem of calculus, we have

$I = f(3) -f(2)$

$= \frac{1}{2}\left \{ \log\left | \frac{3-1}{3+1} \right | - \log\left | \frac{2-1}{2+1} \right | \right \}$

$= \frac{1}{2}\left \{ \log\left | \frac{2}{4} \right | -\log\left | \frac{1}{3} \right | \right \}$

$= \frac{1}{2}\left \{ \log \frac{1}{2} -\log \frac{1}{3} \right \} = \frac{1}{2}\log\frac{3}{2}$

Question:12 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{2}\cos^2 x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{2}\cos^2 x dx$

Consider the integral $\int \cos^2 x dx$

$\int \cos^2 x dx = \int \frac{1+\cos 2x}{2} dx = \frac{x}{2}+\frac{\sin 2x }{4}$

$= \frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{2}) -f(0)$

$= \frac{1}{2}\left \{ \left ( \frac{\pi}{2}-\frac{\sin \pi}{2} \right ) -\left ( 0+\frac{\sin 0}{2} \right ) \right \}$

$= \frac{1}{2}\left \{ \frac{\pi}{2}+0-0-0 \right \}$

$= \frac{\pi}{4}$

Question:13 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3\frac{xdx}{x^2+1}$

Answer:

Given integral: $\int_2^3\frac{xdx}{x^2+1}$

Consider the integral $\int \frac{xdx}{x^2+1}$

$\int \frac{xdx}{x^2+1} = \frac{1}{2}\int \frac{2x}{x^2+1}dx =\frac{1}{2}\log(1+x^2)$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\log(1+x^2)$

Now, by Second fundamental theorem of calculus, we have

$I = f(3) -f(2)$

$= \frac{1}{2}\left \{ \log(1+(3)^2)-\log(1+(2)^2) \right \}$

$= \frac{1}{2}\left \{ \log(10)-\log(5) \right \} = \frac{1}{2}\log\left ( \frac{10}{5} \right ) = \frac{1}{2}\log2$

Question:14 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{2x+3}{5x^2+1}dx$

Answer:

Given integral: $\int_0^1\frac{2x+3}{5x^2+1}dx$

Consider the integral $\int \frac{2x+3}{5x^2+1}dx$

Multiplying by 5 both in numerator and denominator:

$\int \frac{2x+3}{5x^2+1}dx = \frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx$

$=\frac{1}{5}\int \frac{10x+15}{5x^2+1}dx$

$= \frac{1}{5} \int \frac{10x}{5x^2+1} dx +3\int \frac{1}{5x^2+1} dx$

$= \frac{1}{5}\int \frac{10x}{5x^2+1}+3\int \frac{1}{5\left ( x^2+\frac{1}{5} \right )}dx$

$= \frac{1}{5}\log(5x^2+1) +\frac{3}{5}\times \frac{1}{\frac{1}{\sqrt5}} \tan^{-1}\frac{x}{\frac{1}{\sqrt5}}$

$= \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )$

So, we have the function of $x$ , $f(x) = \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \left \{ \frac{1}{5}\log(1+5)+\frac{3}{\sqrt5}\tan^{-1}(\sqrt5) \right \} - \left \{ \frac{1}{5}\log(1)+\frac{3}{\sqrt5}\tan^{-1}(0) \right \}$

$= \frac{1}{5}\log 6 +\frac{3}{\sqrt 5}\tan^{-1}{\sqrt5}$

Question:15 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1xe^{x^2}dx$

Answer:

Given integral: $\int_0^1xe^{x^2}dx$

Consider the integral $\int xe^{x^2}dx$

Putting $x^2 = t$ which gives, $2xdx =dt$

As, $x\rightarrow0 ,t \rightarrow0$ and as $x\rightarrow1 ,t \rightarrow1$ .

So, we have now:

$\therefore I = \frac{1}{2}\int_0^1 e^t dt$

$= \frac{1}{2}\int e^t dt = \frac{1}{2} e^t$

So, we have the function of $x$ , $f(x) = \frac{1}{2} e^t$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \frac{1}{2}e^1 -\frac{1}{2}e^0 = \frac{1}{2}(e-1)$

Question:16 Evaluate the definite integrals in Exercises 1 to 20.

$\int_1^2\frac{5x^2}{x^2 + 4x +3}$

Answer:

Given integral: $I = \int_1^2\frac{5x^2}{x^2 + 4x +3}$

So, we can rewrite the integral as;

$I = \int_1^2 \frac{5x^2}{x^2 + 4x +3}= \int_1^2 \left ( 5 - \frac{20x+15}{x^2 + 4x +3} \right ) dx$

$= \int_1^2 5 dx - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$= [5x]_1^2 - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$I = 5-I_1$ where $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$ . ................(1)

Now, consider $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

Take numerator $20x+15 = A \frac{d}{dx}\left ( x^2+4x+3 \right )+B$

$= 2A x+(4A+B)$

We now equate the coefficients of x and constant term, we get

$A= 10 \and\ B =-25$

$\Rightarrow I_1 = 10\int_1^2 \frac{2x+4}{x^2+4x+3}dx -25\int_1^2 \frac{dx}{x^2+4x+3}$

Now take denominator $x^2+4x+3 = t$

Then we have $(2x+4)dx =dt$

$\Rightarrow I_{1} =10\int \frac{dt}{t} -25\int \frac{dx}{(x+2)^2-1^2}$

$= 10\log t -25\left [ \frac{1}{2}\log\left ( \frac{x+2-1}{x+2+1} \right ) \right ]$

$=[10\log(x^2+4x+3)]_1^2 -25 \left [ \frac{1}{2}\log\left ( \frac{x+1}{x+3} \right ) \right ]_1^2$

$= \left [ 10\log15 -10\log 8 \right ] -25 \left [ \frac{1}{2}\log\frac{3}{5} -\frac{1}{2}\log\frac{2}{4} \right ]$

$= \left [ 10\log5 +10\log3 -10\log4-10\log2 \right ] -\frac{25}{2}\left [ \log3 -\log5-\log2+\log4 \right ]$ $= \left ( 10+\frac{25}{2} \right )\log5 + \left ( -10-\frac{25}{2} \right )\log 4 + \left ( 10-\frac{25}{2} \right )\log 3 + \left ( -10+\frac{25}{2} \right )\log 2$ $= \frac{45}{2}\log5 -\frac{45}{2}\log4 - \frac{5}{2}\log3 +\frac{5}{2}\log2$

$= \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log \frac{3}{2}$

Then substituting the value of $I_{1}$ in equation (1), we get

$I= 5 -\left ( \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2} \right )$

$= 5 -\frac{5}{2}\left ( 9\log\frac{5}{4}-\log\frac{3}{2} \right )$

Question:17 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Answer:

Given integral: $\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Consider the integral $\int (2\sec^2x + x^3 + 2)dx$

$\int (2\sec^2x + x^3 + 2)dx = 2\tan x +\frac{x^4}{4}+2x$

So, we have the function of $x$ , $f(x) = 2\tan x +\frac{x^4}{4}+2x$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(0)$

$= \left \{ \left ( 2\tan\frac{\pi}{4}+\frac{1}{4}\left ( \frac{\pi}{4} \right )^4+2\frac{\pi}{4} \right ) - \left ( 2\tan 0 +0 +0 \right ) \right \}$

$=2\tan\frac{\pi}{4} +\frac{\pi^4}{4^5} +\frac{\pi}{2}$

$2+\frac{\pi}{2}+\frac{\pi^4}{1024}$

Question:18 Evaluate the definite integrals in Exercises 1 to 20.

$\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Answer:

Given integral: $\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Consider the integral $\int (\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

can be rewritten as: $-\int (\cos^2\frac{x}{2} - \sin^2\frac{x}{2})dx = -\int_0^{\pi} \cos x dx$

$= \sin x$

So, we have the function of $x$ , $f(x) =\sin x$

Now, by Second fundamental theorem of calculus, we have

$I = f(\pi) - f(0)$

$\Rightarrow \sin \pi - \sin 0 = 0-0 =0$

Question:19 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^2\frac{6x+3}{x^2+ 4}$

Answer:

Given integral: $\int_0^2\frac{6x+3}{x^2+ 4}$

Consider the integral $\int \frac{6x+3}{x^2+ 4}$

can be rewritten as: $\int \frac{6x+3}{x^2+ 4} = 3\int \frac{2x+1}{x^2+4}dx$

$= 3\int \frac{2x}{x^2+4}dx +3\int \frac{1}{x^2+4}dx$

$= 3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}$

So, we have the function of $x$ , $f(x) =3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(2) - f(0)$

$= \left \{ 3\log(2^2+4)+\frac{3}{2}\tan^{-1}\left ( \frac{2}{2} \right ) \right \}- \left \{ 3\log(0+4)+\frac{3}{2}\tan^{-1}\left ( \frac{0}{2} \right ) \right \}$ $=3\log 8 +\frac{3}{2}\tan^{-1}1 -3\log 4 -\frac{3}{2}\tan^{-1} 0$

$=3\log 8 +\frac{3}{2}\times\frac{\pi}{4} -3\log 4 -0$

$=3\log \frac{8}{4} +\frac{3\pi}{8}$

or we have $=3\log 2 +\frac{3\pi}{8}$

Question:20 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1(xe^x + sin\frac{\pi x}{4})dx$

Answer:

Given integral: $\int_0^1(xe^x + sin\frac{\pi x}{4})dx$

Consider the integral $\int (xe^x + sin\frac{\pi x}{4})dx$

can be rewritten as: $x\int e^x dx - \int \left \{ \left ( \frac{d}{dx}x \right )\int e^x dx \right \}dx +\left \{ \frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}} \right \}$

$= xe^x -\int e^x dx -\frac{4\pi}{\pi} \cos \frac{x}{4}$

$= xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}$

So, we have the function of $x$ , $f(x) = xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) - f(0)$

$= \left (1.e^t-e^t - \frac{4}{\pi}\cos \frac{\pi}{4} \right ) - \left ( 0.e^0 -e^0 -\frac{4}{\pi}\cos 0 \right )$

$= e-e -\frac{4}{\pi}\left ( \frac{1}{\sqrt2} \right )+1+\frac{4}{\pi}$

Question:21 Choose the correct answer in Exercises 20 and 21.

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}$

(A) $\frac{\pi}{3}$

(B) $\frac{2\pi}{3}$

(D) $\frac{\pi}{12}$

Answer:

Given definite integral $\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}$

Consider $\int \frac{dx}{1 +x^2} = \tan^{-1}x$

we have then the function of x, as $f(x) = \tan^{-1}x$

By applying the second fundamental theorem of calculus, we will get

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2} = f(\sqrt3) - f(1)$

$= \tan^{-1}\sqrt{3} - \tan^{-1}1$

$=\frac{\pi}{3} - \frac{\pi}{4}$

$= \frac{\pi}{12}$

Therefore the correct answer is D.

Question:22 Choose the correct answer in Exercises 21 and 22.

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}$ equals

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{12}$

(D) $\frac{\pi}{4}$

Answer:

Given definite integral $\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}$

Consider $\int \frac{dx}{4+ 9x^2} = \int \frac{dx}{2^2+(3x)^2}$

Now, putting $3x = t$

we get, $3dx=dt$

Therefore we have, $\int \frac{dx}{2^2+(3x)^2} = \frac{1}{3}\int \frac{dt}{2^2+t^2}$

$= \frac{1}{3}\left ( \frac{1}{2}\tan^{-1}\frac{t}{2} \right ) = \frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )$

we have the function of x , as $f(x) =\frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )$

So, by applying the second fundamental theorem of calculus, we get

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} = f(\frac{2}{3}) - f(0)$

$= \frac{1}{6}\tan^{-1}\left ( \frac{3}{2}.\frac{2}{3} \right ) -\frac{1}{6}\tan^{-1}0$

$= \frac{1}{6}\tan^{-1}1 - 0$

$= \frac{1}{6}\times \frac{\pi}{4} = \frac{\pi}{24}$

Therefore the correct answer is C.

NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.10

Question:1 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^1\frac{x}{x^2 +1}dx$

Answer:

$\int_0^1\frac{x}{x^2 +1}dx$
let $x^2+1 = t \Rightarrow xdx =dt/2$
when x = 0 then t = 1 and when x =1 then t = 2
$\therefore \int_{o}^{1}\frac{x}{x^2+1}dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}$
$\\=\frac{1}{2}[\log\left | t \right |]_{1}^{2}\\ =\frac{1}{2}\log 2$

Question:2 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$

Answer:

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$
let $\sin \phi = t \Rightarrow \cos \phi d\phi = dt$
when $\phi =0,t\rightarrow 0$ and $\phi =\pi/2,t\rightarrow 1$

using the above substitution we can evaluate the integral as

$\\\therefore \int_{0}^{1}\sqrt{t}(1-t^2)dt\\ =\int_{0}^{1} t^\frac{1}{2}(1+t^4-2t^2)dt\\ =\int_{0}^{1}t^\frac{1}{2}dt+\int_{0}^{1}t^{9/2}dt-2\int_{0}^{1}t^{5/2}dt\\ =[2t^{3/2}/3+2t^{11/2}/11+4t^{7/2}/7]^1_0\\ =\frac{64}{231}$

Question:3 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$

Answer:

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
let $x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta$
when x = 0 then $\theta= 0$ and when x = 1 then $\theta= \pi/4$

$\\=\int_{0}^{\pi/4}\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}2\theta \sec^2\theta d\theta\\$
Taking $\theta$ as a first function and $\sec^2\theta$ as a second function, by using by parts method

$\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]^{\pi/4}_0\\ =2[\pi/4+\log(1/\sqrt{2})]\\ =\pi/4-\log 2$

Question:4 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^2x\sqrt{x+2}$ . (Put ${x+2} = t^2$ )

Answer:

Let $x+2 = t^2\Rightarrow dx =2tdt$
when x = 0 then t = $\sqrt{2}$ and when x=2 then t = 2

$I=\int_{0}^{2}x\sqrt{x+2}dx$

$\\=2\int_{\sqrt{2}}^{2}(t^2-2)t^2dt\\ =2\int_{\sqrt{2}}^{2}(t^4-2t^2)dt\\ =2[t^5/5-\frac{2}{3}t^3]^2_{\sqrt{2}}\\ =2[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}]\\ =\frac{16\sqrt{2}(\sqrt{2}+1)}{15}$

Question:5 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx$

Answer:

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx =I$
let $\cos x =t\Rightarrow -\sin x dx = dt$
when x=0 then t = 1 and when x= $\pi/2$ then t = 0

$\\I=\int_{1}^{0}\frac{dt}{1+t^2}\\ =[\tan ^{-1}t]^0_1\\ =\pi/4$

Question:6 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^2\frac{dx}{x + 4 - x^2}$

Answer:

By adjusting, the denominator can also be written as $(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2 =x+4-x^2$
Now,
$\Rightarrow \int_{0}^{2}\frac{dx}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}$
let $x-1/2 = t\Rightarrow dx=dt$
when x= 0 then t =-1/2 and when x =2 then t = 3/2

$\\\Rightarrow\int_{-1/2}^{3/2}\frac{dt}{(\frac{\sqrt{17}}{2})^2-t^2}\\ =\frac{1}{2.\frac{\sqrt{17}}{2}}\log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}/2+3/2}{\sqrt{17}/2-3/2}-\log\frac{\sqrt{17}/2-1/2}{\sqrt{17}/2+1/2}]\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}+3}{\sqrt{17}-3/}.\frac{\sqrt{17}+1}{\sqrt{17}+1}]$
$\\ =\frac{1}{\sqrt{17}}[\log (\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}})]\\ =\frac{1}{\sqrt{17}}[\log (\frac{5+\sqrt{17}}{5-\sqrt{17}})]$
On rationalisation, we get

$=\frac{1}{\sqrt{17}}\log \frac{21+5\sqrt{17}}{4}$

Question:7 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$

Answer:

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$
the Dr can be written as $x^2+2x+5 = (x+1)^2+2^2$
and put x+1 = t then dx =dt

when x= -1 then t = 0 and when x = 1 then t = 2

$\\\Rightarrow \int_{0}^{2}\frac{dt}{t^2+2^2}\\ =\frac{1}{2}[\tan^{-1}\frac{t}{2}]^2_0\\ =\frac{1}{2}( \pi/4)\\ =\frac{\pi}{8}$

Question:8 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$

Answer:

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$
let $2x =t \Rightarrow 2dx =dt$
when x = 1 then t = 2 and when x = 2 then t= 4

$\\=\frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{2}{t^2})e^tdt\\$
let
$\frac{1}{t} = f(t)\Rightarrow f'(t)=-\frac{1}{t^2}$
$\Rightarrow \int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^tdt =\int_{2}^{}4e^t[f(t)+f'(t)]dt$
$\\=[e^tf(t)]^4_2\\ =[e^t.\frac{1}{t}]^4_2\\ =\frac{e^4}{4}-\frac{e^2}{2}\\ =\frac{e^2(e^2-2)}{4}$

Question:9 Choose the correct answer in Exercises 9 and 10.

The value of the integral $\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$ is

(A) 6

(B) 0

(D) 4

Answer:

The value of integral is (A) = 6

$\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$
$\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^\frac{1}{3}}{x^3}dx\\$
let
$\frac{1}{x^2}-1 = t\Rightarrow \frac{dx}{x^3}=-dt/2$
now, when x = 1/3, t = 8 and when x = 1 , t = 0

therefore

$\\=-\frac{1}{2}\int_{8}^{0}t^{1/3}dt\\ =-\frac{1}{2}.\frac{3}{4}[t^4/3]^0_8\\ =-\frac{3}{8}[-2^4]\\ =6$

Question:10 Choose the correct answer in Exercises 9 and 10.

If $f(x) = \int_0^x t \sin t dt$ , then $f'(x)$ is

(A) $\cos x + x\sin x$

(B) $x\sin x$

(D) $\sin x + x\cos x$

Answer:

The correct answer is (B) = $x\sin x$

$f(x) = \int_0^x t \sin t dt$
by using by parts method,
$\\=t\int \sin t dt - \int (\frac{d}{dt}t\int \sin t dt)dt\\ =[t(-\cos t )+\sin t]^x_0$

$f(x)= -x\cos x+sinx$
So, $f'(x)= -\cos x+x\sin x+\cos x\\ =x\sin x$

NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.11

Question:1 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{2}\cos^2 x dx$

Answer:

We have _{.............................................................} (i)

By using

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get :-

$I\ =\ \int_0^\frac{\pi}{2}\sin^2 x dx$ ................................................................ (ii)

Adding both (i) and (ii), we get :-

$\int_0^\frac{\pi}{2}\cos^2 x dx$ $+\ \int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2I$

or $\int_0^\frac{\pi}{2}\ (cos^2 x\ +\ sin^2 x) dx\ =\ 2I$

or $\int_0^\frac{\pi}{2}1. dx\ =\ 2I$

or $2I\ =\ \left [ x \right ] ^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}$

or $I\ =\ \frac{\Pi }{4}$

Question:2 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. $\int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx$

Answer:

We have _{..........................................................................} (i)

By using ,

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

or $I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx$ .......................................................(ii)

Adding (i) and (ii), we get,

$2I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}\ +\ \sqrt{\cos x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx$

or $2I\ =\ \int_0^\frac{\pi}{2}1.dx$

or $2I\ =\ \left [ x \right ]^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}$

Thus $I\ =\ \frac{\Pi }{4}$

Question:3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$

Answer:

We have $I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$ ..................................................................(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x)dx}{\sin^\frac{3}{2}(\frac{\pi}{2}-x) + \cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}$

or $I\ =\ \int^{\frac{\pi}{2}}_0\frac{\cos^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$ . ............................................................(ii)

Adding (i) and (ii), we get :

$2I\ =\ \int^{\frac{\pi}{2}}_0\frac{\ (sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x)dx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$

or $2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx$

or $2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}$

Thus $I\ =\ {\frac{\pi}{4}}$

Question:4 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. $\int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}$ ..................................................................(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 (\frac{\pi}{2}-x)dx}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)}$

or $I\ =\ \int_0^\frac{\pi}{2} \frac{\sin^5 xdx}{\sin^5x + \cos^5x}$ . ............................................................(ii)

Adding (i) and (ii), we get :

$2I\ =\ \int_0^\frac{\pi}{2} \frac{\ ( sin^5x \ +\ cos^5 x)dx}{\sin^5x + \cos^5x}$

or $2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx$

or $2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}$

Thus $I\ =\ {\frac{\pi}{4}}$

Question:5 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_{-5}^5|x+2|dx$

Answer:

We have, $I\ =\ \int_{-5}^5|x+2|dx$

For opening the modulas we need to define the bracket :

If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-

$I\ =\ \int_{-5}^{-2} -(x+2)dx\ +\ \int_{-2}^{5} (x+2)dx$

or $I\ =\ -\left [ \frac{x^2}{2}\ +\ 2x \right ]^{-2} _{-5}\ +\ \left [ \frac{x^2}{2}\ +\ 2x \right ]^{5} _{-2}$

This gives $I\ =\ 29$

Question:6 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_2^8|x-5|dx$

Answer:

We have, $I\ =\ \int_{2}^8|x-5|dx$

For opening the modulas we need to define the bracket :

If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-

$I\ =\ \int_{2}^{5} -(x-5)dx\ +\ \int_{5}^{8} (x-5)dx$

or $I\ =\ -\left [ \frac{x^2}{2}\ -\ 5x \right ]^{5} _{2}\ +\ \left [ \frac{x^2}{2}\ -\ 5x \right ]^{8} _{5}$

This gives $I\ =\ 9$

Question:7 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int^1_0x(1-x)^ndx$

Answer:

We have $I\ =\ \int^1_0x(1-x)^ndx$

U sing the property : -

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get : -

$I\ =\ \int^1_0x(1-x)^ndx\ =\ \int^1_0(1-x)(1-(1-x))^ndx$

or $I\ =\ \int^1_0(1-x)x^n\ dx$

or $I\ =\ \int^1_0(x^n\ -\ x^{n+1}) \ dx$

or $=\ \left [ \frac{x^{n+1}}{n+1}\ -\ \frac{x^{n+2}}{n+2} \right ]^1_0$

or $=\ \left [ \frac{1}{n+1}\ -\ \frac{1}{n+2} \right ]$

or $I\ =\ \frac{1}{(n+1)(n+2)}$

Question:8 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{4}\log(1+\tan x)dx$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx$

By using the identity

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx\ =\ \int_0^\frac{\pi}{4}\log(1+\tan (\frac{\pi}{4}-x))dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log(1+\frac{1-\tan x}{1+\tan x})dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log(\frac{2}{1+\tan x})dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ \int_0^\frac{\pi}{4}\log(1+ \tan x)dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ I$

or $2I\ =\ \left [ x\log2 \right ]^{\frac{\Pi }{4}}_0$

or $I\ =\ \frac{\Pi }{8}\log2$

Question:9 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^2x\sqrt{2-x}dx$

Answer:

We have $I\ =\ \int_0^2x\sqrt{2-x}dx$

By using the identity

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get :

$I\ =\ \int_0^2x\sqrt{2-x}dx\ =\ \int_0^2(2-x)\sqrt{2-(2-x)}dx$

or $I\ =\ \int_0^2(2-x)\sqrt{x}dx$

or $I\ =\ \int_0^2(2\sqrt{x}\ -\ x^\frac{3}{2} dx$

or $=\ \left [ \frac{4}{3}x^\frac{3}{2}\ -\ \frac{2}{5}x^\frac{5}{2} \right ]^2_0$

or $=\ \frac{4}{3}(2)^\frac{3}{2}\ -\ \frac{2}{5}(2)^\frac{5}{2}$

or $I\ =\ \frac{16\sqrt{2}}{15}$

Question:10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx$

or $I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log(2\sin x\cos x))dx$

or $I\ =\ \int_0^\frac{\pi}{2} (\log\sin x- \log\cos x\ -\ \log2)dx$ ..............................................................(i)

By using the identity :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get :

$I\ =\ \int_0^\frac{\pi}{2} (\log\sin (\frac{\pi}{2}-x)- \log\cos (\frac{\pi}{2}-x)\ -\ \log2)dx$

or $I\ =\ \int_0^\frac{\pi}{2} (\log\cos x- \log\sin x\ -\ \log2)dx$ ....................................................................(ii)

Adding (i) and (ii) we get :-

$2I\ =\ \int_0^\frac{\pi}{2} (- \log 2 -\ \log 2)dx$

or $I\ =\ -\log 2\left [ \frac{\Pi }{2} \right ]$

or $I\ =\ \frac{\Pi }{2}\log\frac{1}{2}$

Question:11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx$

Answer:

We have $I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx$

We know that sin ² x is an even function. i.e., sin ² (-x) = (-sinx) ² = sin ² x.

Also,

$I\ =\ \int_{-a}^af(x) dx\ =\ 2\int_{0}^af(x) dx$

So,

$I\ =\ 2\int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2\int_0^\frac{\pi}{2}\frac{(1-\cos2x)}{2} dx$

or $=\ \left [ x\ -\ \frac{\sin2x}{2} \right ]^{\frac{\Pi }{2}}_0$

or $I\ =\ \frac{\Pi }{2}$

Question:12 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\pi\frac{xdx}{1+\sin x}$

Answer:

We have $I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}$ ..........................................................................(i)

By using the identity :-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin (\Pi -x)}$

or $I\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin x}$ ............................................................................(ii)

Adding both (i) and (ii) we get,

$2I\ =\ \int_0^\pi\frac{\Pi}{1+\sin x} dx$

or $2I\ =\ \Pi \int_0^\pi\frac{1-\sin x}{(1+\sin x)(1-\sin x)} dx\ =\ \Pi \int_0^\pi\frac{1-\sin x}{\cos^2 x} dx$

or $2I\ =\ \Pi \int_0^\pi (\sec^2\ -\ \tan x \sec x) x dx$

or $I\ =\ \Pi$

Question:13 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx$

Answer:

We have $I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx$

We know that $\sin^7x$ is an odd function.

So the following property holds here:-

$\int_{-a}^{a}f(x)dx\ =\ 0$

Hence

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx\ =\ 0$

Question:14 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^{2\pi}\cos^5xdx$

Answer:

We have $I\ =\ \int_0^{2\pi}\cos^5xdx$

I t is known that :-

$\int_0^{2a}f(x)dx\ =\ 2\int_0^{a}f(x)dx$ If f (2a - x) = f(x)

$=\ 0$ If f (2a - x) = - f(x)

Now, using the above property

$\cos^5(\Pi - x)\ =\ - \cos^5x$

Therefore, $I\ =\ 0$

Question:15 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx$

Answer:

We have $I\ =\ \int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx$ ................................................................(i)

By using the property :-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get ,

$I\ =\ \int^\frac{\pi}{2} _0\frac{\sin (\frac{\pi}{2}-x) - \cos (\frac{\pi}{2}-x) }{1+\sin (\frac{\pi}{2}-x)\cos (\frac{\pi}{2}-x)}dx$

or $I\ =\ \int^\frac{\pi}{2} _0\frac{\cos x - \sin x }{1+\sin x\cos x}dx$ ......................................................................(ii)

Adding both (i) and (ii), we get

$2I\ =\ \int^\frac{\pi}{2} _0\frac{0 }{1+\sin x\cos x}dx$

Thus I = 0

Question:16 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\pi\log(1 +\cos x)dx$

Answer:

We have $I\ =\ \int_0^\pi\log(1 +\tan x)dx$ .....................................................................................(i)

By using the property:-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\pi\log(1 +\cos (\Pi -x))dx$

$I\ =\ \int_0^\pi\log(1 -\cos x)dx$ ....................................................................(ii)

Adding both (i) and (ii) we get,

$2I\ =\ \int_0^\pi\log(1 +\cos x)dx\ +\ \int_0^\pi\log(1 -\cos x)dx$

or $2I\ =\ \int_0^\pi\log(1 -\cos^2 x)dx\ =\ \int_0^\pi\log \sin^2 xdx$

or $2I\ =\ 2\int_0^\pi\log \sin xdx$

or $I\ =\ \int_0^\pi\log \sin xdx$ ........................................................................(iii)

or $I\ =\ 2\int_0^ \frac{\pi}{2} \log \sin xdx$ ........................................................................(iv)

or $I\ =\ 2\int_0^ \frac{\pi}{2} \log \cos xdx$ .....................................................................(v)

Adding (iv) and (v) we get,

$I\ =\ -\pi \log2$

Question:17 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx$

Answer:

We have $I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx$ ................................................................................(i)

By using, we get

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx\ =\ \int_0^a \frac{\sqrt {(a-x)}}{\sqrt {(a-x)} + \sqrt{x}}dx$ .................................................................(ii)

Adding (i) and (ii) we get :

$2I\ =\ \int_0^a \frac{\sqrt x\ +\ \sqrt{a-x}}{\sqrt x + \sqrt{a-x}}dx$

or $2I\ =\ \left [ x \right ]^a_0 = a$

or $I\ =\ \frac{a}{2}$

Question:18 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^4 |x-1|dx$

Answer:

We have, $I\ =\ \int_{0}^4|x-1|dx$

For opening the modulas we need to define the bracket :

If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-

$I\ =\ \int_{0}^{1} -(x-1)dx\ +\ \int_{1}^{4} (x-1)dx$

or $I\ =\ \left [ x\ -\ \frac{x^2}{2}\ \right ]^{1} _{0}\ +\ \left [ \frac{x^2}{2}\ -\ x \right ]^{4} _{1}$

This gives $I\ =\ 5$

Question:19 Show that $\int_0^a f(x)g(x)dx = 2\int_0^af(x)dx$ if $f$ and $g$ are defined as $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$

Answer:

Let $I\ =\ \int_0^a f(x)g(x)dx$ ........................................................(i)

This can also be written as :

$I\ =\ \int_0^a f(a-x)g(a-x)dx$

or $I\ =\ \int_0^a f(x)g(a-x)dx$ ................................................................(ii)

Adding (i) and (ii), we get,

$2I\ =\ \int_0^a f(x)g(a-x)dx +\ \int_0^a f(x)g(x)dx$

$2I\ =\ \int_0^a f(x)4dx$

or $I\ =\ 2\int_0^a f(x)dx$

Question:20 Choose the correct answer in Exercises 20 and 21.

The value of is $\int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx$ is

(A) 0

(B) 2

(D) 1

Answer:

We have

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx$

This can be written as :

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}x^3dx +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} x\cos x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} \tan^5 x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} 1dx$

Also if a function is even function then $\int_{-a}^{a}f(x)\ dx\ =\ 2\int_{0}^{a}f(x)\ dx$

And if the function is an odd function then : $\int_{-a}^{a}f(x)\ dx\ =\ 0$

Using the above property I become:-

$I\ =\ 0+0+0+ 2\int_{0}^{\frac{\Pi }{2}}1.dx$

or $I\ =\ 2\left [ x \right ]^{\frac{\Pi }{2}}_0$

or $I\ =\ \Pi$

Question:21 Choose the correct answer in Exercises 20 and 21.

The value of $\int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx$ is

Answer:

We have

$I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx$ .................................................................................(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin (\frac{\pi}{2}-x)}{4+3\cos (\frac{\pi}{2}-x)} \right )dx$

or $I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx$ .............................................................................(ii)

Adding (i) and (ii), we get:

$2I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ +\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx$

or $2I\ =\ \int_0^\frac{\pi}{2}\log1.dx$

Thus $I\ =\ 0$

NCERT solutions for class 12 maths chapter 7 Integrals-Miscellaneous Exercise

Question:1 Integrate the functions in Exercises 1 to 24.

$\frac{1}{x - x^3}$

Answer:

Firstly we will simplify the given equation :-

$\frac{1}{x - x^3}\ =\ \frac{1}{(x)(1-x)(1+x)}$

Let

$\frac{1}{(x)(1-x)(1+x)} =\ \frac{A}{x}\ +\ \frac{B}{1-x}\ +\ \frac{C}{1+x}$

By solving the equation and equating the coefficients of x ² , x and the constant term, we get

$A\ =\ 1,\ B\ =\ \frac{1}{2},\ C\ =\ \frac{-1}{2}$

Thus the integral can be written as :

$\int \frac{1}{(x)(1-x)(1+x)}dx =\ \int \frac{1}{x}dx\ +\ \frac{1}{2}\int \frac{1}{1-x}dx\ +\ \frac{-1}{2}\int \frac{1}{1+x}dx$

$=\ \log x\ -\ \frac{1}{2}\log(1-x)\ +\ \frac{-1}{2}\log (1+x)$

or $=\ \frac{1}{2} \log \frac{x^2}{1-x^2}\ +\ C$

Question:2 Integrate the functions in Exercises 1 to 24.

$\frac{1}{\sqrt{x+a} + \sqrt{x+b}}$

Answer:

At first we will simplify the given expression,

$\frac{1}{\sqrt{x+a} + \sqrt{x+b}}\ =\ \frac{1}{\sqrt{x+a} + \sqrt{x+b}}\times\frac{\sqrt{x+a} - \sqrt{x+b}}{\sqrt{x+a} - \sqrt{x+b}}$

or $=\ \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b}$

Now taking its integral we get,

$\int \frac{1}{\sqrt{x+a} + \sqrt{x+b}}\ =\ \frac{1}{a-b}\int (\sqrt{x+a} -\sqrt{x+b})dx$

or $=\ \frac{1}{a-b}\left [ \frac{(x+a)^{\frac{3}{2}}} {\frac{3}{2}}\ -\ \frac{(x+b)^{\frac{3}{2}}} {\frac{3}{2}} \right ]$

or $=\ \frac{2}{3(a-b)}\left [ (x+a)^{\frac{3}{2}}\ -\ (x+b)^{\frac{3}{2}} \right ]\ +\ C$

Question:3 Integrate the functions in Exercises 1 to 24.

$\frac{1}{x\sqrt{ax-x^2}}$ [Hint: Put $x = \frac{a}{t}$ ]

Answer:

Let

$x = \frac{a}{t}\ dx\ \Rightarrow \ dx\ =\ \frac{-a}{t^2}dh$

Using the above substitution we can write the integral is

$\int \frac{1}{x\sqrt{ax-x^2}}\ =\ \int \frac{1}{\frac{a}{t}\sqrt{a.\frac{a}{t}\ -\ (\frac{a}{t})^2}} \frac{-a}{t^2}dt$

$=\ \frac{-1}{a}\int \frac{1}{\sqrt{(t-1)}}dt$

$=\ \frac{-1}{a}\ (2\sqrt{t-1})\ +\ C$

or $=\ \frac{-1}{a}\ (2\sqrt{\frac{a}{x}\ -\ 1})\ +\ C$

or $=\ \frac{-2}{a}\ \sqrt{\frac{a-x}{x}}\ +\ C$

Question:4 Integrate the functions in Exercises 1 to 24.

$\frac{1}{x^2(x^4 + 1)^\frac{3}{4}}$

Answer:

For the simplifying the expression, we will multiply and dividing it by x ^-3 .

We then have,

$\frac{x^{-3}}{x^2 x^{-3}(x^4 + 1)^\frac{3}{4}}\ =\ \frac{1}{x^5}\left [ \frac{x^4\ +\ 1}{x^4} \right ]^{\frac{-3}{4}}$

Now, let

$\frac{1}{x^4}\ =\ t\ \Rightarrow \ \frac{1}{x^5}dx\ =\ \frac{-dt}{4}$

Thus,

$\int \frac{1}{x^2(x^4 + 1)^\frac{3}{4}}\ =\ \int \frac{1}{x^5}\left ( 1+\ \frac{1}{x^4}^{\frac{-3}{4}}\ \right )dx$

or $=\ \frac{-1}{4} \int (1+t)^{\frac{-3}{4}}dt$

$=\ \frac{-1}{4} \frac{(1+\frac{1}{x^4})^{\frac{1}{4}}}{\frac{1}{4}}\ +\ C$

$=\ - \left [ 1+\frac{1}{x^4} \right ]^{\frac{1}{4}}\ +\ C$

Question:5 Integrate the functions in Exercises 1 to 24.

$\frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}}$ [Hint: $\frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}} = \frac{1}{x^\frac{1}{3}(1 + \ x^\frac{1}{6})}$ , put $x = t^6$ ]

Answer:

Put $x = t^6\ \Rightarrow \ dx = 6t^5dt$

We get,

$\int \frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}}dx\ =\ \int \frac{6t^5}{t^3+t^2}dt$

or $=\ 6\int \frac{t^3}{1+t}dt$

or $=\ 6\int \left \{ (t^2-t+1)-\frac{1}{1+t} \right \}dt$

or $=\ 6 \left [ \left ( \frac{t^3}{3} \right ) -\left ( \frac{t^2}{2} \right )+t - \log(1+t) \right ]$

Now put $x = t^6$ in the above result :

$=\ 2\sqrt{x} -3x^{\frac{1}{3}}+ 6x^{\frac{1}{6}} - 6 \log \left ( 1-x^\frac{1}{6} \right )\ +\ C$

Question:6 Integrate the functions in Exercises 1 to 24.

$\frac{5x}{(x+1)(x^2 + 9)}$

Answer:

Let us assume that :

$\frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{A}{(x+1)}\ +\ \frac{Bx + c}{x^2 + 9}$

Solving the equation and comparing coefficients of x ² , x and the constant term.

We get,

$A\ =\ \frac{-1}{2}\ ;\ B\ =\ \frac{1}{2}\ ;\ C\ =\ \frac{9}{2}$

Thus the equation becomes :

$\frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{-1}{2(x+1)}\ +\ \frac{\frac{x}{2}+\frac{9}{2}}{x^2 + 9}$

$\int \frac{5x}{(x+1)(x^2 + 9)}\ =\ \int \left [ \frac{-1}{2(x+1)}\ +\ \frac{x+9}{2(x^2 + 9}) \right ]dx$

or $=\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{2} \int \frac{x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx$

or $=\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \int \frac{2x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx$

or $=\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \log {(x^2 +9)} +\frac{3}{2} \tan^{-1}\frac{x}{3}\ +\ C$

Question:7 Integrate the functions in Exercises 1 to 24.

$\frac{\sin x}{\sin (x-a)}$

Answer:

We have,

$I\ =\ \frac{\sin x}{\sin (x-a)}$

Assume :- $(x-a)\ =\ t \Rightarrow \ dx=dt$

Putting this in above integral :

$\int \frac{\sin x}{\sin (x-a)}dx\ =\ \int \frac{\sin (t+a)}{\sin t}dt$

or $=\ \int \frac{\sin t \cos a\ +\ \cos t \sin a }{\sin t}dt$

or $=\ \int (\cos a\ +\ \cot t \sin a)dt$

or $=\ t\cos a\ +\ \sin a \log |\sin t|\ +\ C$

or $=\ \sin a \log \left | \sin(x-a) \right | + x\cos a\ +\ C$

Question:9 Integrate the functions in Exercises 1 to 24.

$\frac{\cos x}{\sqrt{4 - \sin^2 x}}$

Answer:

We have the given integral

$I\ =\ \frac{\cos x}{\sqrt{4 - \sin^2 x}}$

Assume $\sin x = t\ \Rightarrow \cos x dx = dt$

So, this substitution gives,

$\int \frac{\cos x}{\sqrt{4 - \sin^2 x}}\ =\ \int \frac{dt}{\sqrt{(2)^2 - (t)^2}}$

$=\ \sin^{-1}\frac{t}{2}\ +\ C$

or $=\ \sin^{-1}\left ( \frac{\sin x}{2} \right )\ +\ C$

Question:10 Integrate the functions in Exercises 1 to 24.

$\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}$

Answer:

We have

$I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}$

Simplifying the given expression, we get :

$\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ \frac{(\sin^4x + \cos^4x)(\sin^4x - \cos^4x) }{1- 2\sin^ x\cos^2 x}$

or $=\ \frac{(\sin^4x + \cos^4x)(\sin^2x - \cos^2x)(\sin^2x + \cos^2x) }{1- 2\sin^ x\cos^2 x}$

or $=\ -\frac{(\sin^4x + \cos^4x)(\cos^2x - \sin^2x) }{1- 2\sin^ x\cos^2 x}$

or $=\ -\cos^2x - \sin^2x\ =\ -\cos 2x$

Thus,

$I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ -\int \cos 2x\ dx$

and $=\ - \frac{\sin 2x}{2}\ +\ C$

Question:11 Integrate the functions in Exercises 1 to 24.

$\frac{1}{\cos(x+a)\cos(x+b)}$

Answer:

For simplifying the given equation, we need to multiply and divide the expression by _.

Thus we obtain :

$\frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin(a-b)}\times\frac{\sin (a-b)}{\cos(x+a)\cos(x+b)}$

or $= \frac{1}{ \sin (a-b)}\times \frac{\sin{\left [ (x+a) - (x+b) \right ]}}{\cos (x+a) \cos (x+b)}$

or $= \frac{1}{ \sin (a-b)}\times \left ( \frac{\sin (x+a) }{\cos (x+a) } - \frac{\sin(x+b)}{\cos (x+b)} \right )$

or $= \frac{1}{ \sin (a-b)}\times \left ( \tan(x+a)\ -\ \tan(x+b) \right )$

Thus integral becomes :

$\int \frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin (a-b)} \times \int \left ( \tan(x+a)\ -\ \tan(x+b) \right )dx$

or $=\ \frac{1}{\sin (a-b)} \times \left [ -\log \left | \cos (x+a) \right | + \log \left | \cos(x+b) \right | \right ]\ +\ C$

or $=\ \frac{1}{\sin (a-b)} \times \log \left [ \frac{\cos(x+b) }{cos(x+a)} \right ]\ +\ C$

Question:12 Integrate the functions in Exercises 1 to 24.

$\frac{x^3}{\sqrt{1-x^8}}$

Answer:

Given that to integrate

$\frac{x^3}{\sqrt{1-x^8}}$

Let $x^4 = t \implies 4x^3dx = dt$

$\therefore \int \frac{x^3}{\sqrt{1-x^8}}dx = \frac{1}{4}\int\frac{1}{\sqrt {1-t^2}}dt$

$= \frac{1}{4}sin^{-1}t + C= \frac{1}{4}sin^{-1}{x^4} + C$

the required solution is $\frac{1}{4}sin^{-1}{(x^4)} + C$

Question:13 Integrate the functions in Exercises 1 to 24.

$\frac{e^x}{(1 + e^x)(2 + e^x)}$

Answer:

we have to integrate the following function

$\frac{e^x}{(1 + e^x)(2 + e^x)}$

Let $1+e^x = t \implies e^xdx = dt$

using this we can write the integral as

$\therefore \int\frac{e^x}{(1 + e^x)(2 + e^x)}dx = \int\frac{1}{t(1+t)}dt = \int\frac{(1+t)-t}{t(1+t)}dt$

$\\ = \int\left ( \frac{1}{t}-\frac{1}{t+1} \right )dt$

$\\ = \int\frac{1}{t}dt - \int\frac{1}{t+1}dt$

$\\ = \log t - \log (1+t) + C \\ = \log (1+e^x) - \log (2+e^x) + C \\ = \log\left ( \frac{e^x + 1}{e^x + 2} \right ) + C$

Question:14 Integrate the functions in Exercises 1 to 24.

$\frac{1}{(x^2 + 1)(x^2 +4)}$

Answer:

Given,

$\frac{1}{(x^2 + 1)(x^2 +4)}$

Let $I = \int\frac{1}{(x^2 + 1)(x^2 +4)}$

Now, Using partial differentiation,

$\frac{1}{(x^2 + 1)(x^2 +4)} = \frac{Ax + B}{(x^2 + 1)} + \frac{Cx +D}{(x^2 +4)}$

$\implies \frac{1}{(x^2 + 1)(x^2 +4)} = \frac{(Ax + B)(x^2 +4) + (Cx +D)(x^2 + 1)}{(x^2 + 1)(x^2 +4)}$
$\\ \implies1 = (Ax + B)(x^2 + 4)+(Cx + D)(x^2 + 1) \\ \implies 1 = Ax^3 +4Ax+ Bx^2 + 4B+ Cx^3 + Cx + Dx^2 + D \\ \implies (A+C)x^3 +(B+D)x^2 +(4A+C)x + (4B+D) = 1$

Equating the coefficients of $x, x^2, x^3$ and constant value,

A + C = 0 $\implies$ C = -A

B + D = 0 $\implies$ B = -D

4A + C =0 $\implies$ 4A = -C $\implies$ 4A = A $\implies$ A = 0 = C

4B + D = 1 $\implies$ 4B – B = 1 $\implies$ B = 1/3 = -D

Putting these values in equation, we have

155444707519314

155444707593450

1554447076674147

1554447078146237

$\implies I = \frac{1}{3}tan^{-1}x - \frac{1}{6}tan^{-1}\frac{x}{2} + C$

Question:15 Integrate the functions in Exercises 1 to 24.

$\cos^3 x \;e^{\log\sin x}$

Answer:

Given,

$\cos^3 x \;e^{\log\sin x}$

$I = \int \cos^3 x \;e^{\log\sin x}$ (let)

Let $cos x = t \implies -sin x dx = dt \implies sin x dx = -dt$

using the above substitution the integral is written as

$\therefore \int cos^3xe^{\log sinx}dx = \int cos^3x.sinx dx$

155444708344321

1554447084197794

1554447084982414

155444708573581

$I = -\frac{cos^4x}{4} + C$

Question:16Integrate the functions in Exercises 1 to 24.

$e^{3\log x} (x^4 + 1)^{-1}$

Answer:

Given the function to be integrated as

$e^{3\log x} (x^4 + 1)^{-1}$
$= e^{\log x^3}(x^4 + 1)^{-1} = \frac{x^3}{x^4 + 1}$

Let $I = \int e^{3\log x} (x^4 + 1)^{-1}$

Let $x^4 = t \implies 4x^3 dx = dt$

$I = \int e^{3\log x} (x^4 + 1)^{-1} = \int \frac{x^3}{x^4 + 1}$

1554447091784564

1554447092538842

$\implies I = \frac{1}{4}\log(x^4 +1) + C$

Question:17 Integrate the functions in Exercises 1 to 24.

$f'(ax +b)[f(ax +b)]^n$

Answer:

Given,

$f'(ax +b)[f(ax +b)]^n$

Let $I = \int f'(ax +b)[f(ax +b)]^n$

Let f(ax +b) = t ⇒ a .f ' (ax + b)dx = dt

Now we can write the ntegral as

$\int f'(ax +b)[f(ax +b)]^n = \frac{1}{a}\int t^ndt$

$\\ = \frac{1}{a}.\frac{t^{n+1}}{n+1} + C \\ = \frac{1}{a}.\frac{(f(ax+b))^{n+1}}{n+1} + C$

$\implies I = \frac{(f(ax+b))^{n+1}}{a(n+1)} + C$

Question:18 Integrate the functions in Exercises 1 to 24.

$\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$

Answer:

Given,

$\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$

Let $I = \int \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$

We know the identity that

sin (A+B) = sin A cos B + cos A sin B

$\therefore \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} = \frac{1}{\sqrt{\sin^3 x (\sin x \cos \alpha + \cos x \sin \alpha)}}$

$= \frac{1}{\sqrt{\sin^3 x . \sin x(\cos \alpha + \cot x \sin \alpha)}} = \frac{1}{\sqrt{\sin^4 x (\cos \alpha + \cot x \sin \alpha)}}$

$\frac{cosec^2 x}{\sqrt{(\cos \alpha + \cot x \sin \alpha)}}$

1554447105407729

1554447106158144

1554447106897484

1554447107634823

1554447108399148

155444710916564

1554447109907513

1554447110653730

1554447111419992

1554447112159711

Question:19 Integrate the functions in Exercises 1 to 24.

. $\frac{\sin^{-1}\sqrt x - \cos^{-1}\sqrt x}{\sin^{-1}\sqrt x + \cos^{-1}\sqrt x}, \;\;\; x\in [0,1]$

Answer:

We have

$I\ =\ \int \frac{\sin^{-1}\sqrt x - \cos^{-1}\sqrt x}{\sin^{-1}\sqrt x + \cos^{-1}\sqrt x}\ dx$

or $=\ \int \frac{\sin^{-1}\sqrt x - \left ( \frac{\Pi }{2} - \sin^{-1}\sqrt x \right )}{\frac{\Pi }{2}}\ dx$

or $=\ \frac{2}{\Pi } \int \left ( \ 2sin^{-1}\sqrt x - \frac{\Pi }{2} \right )\ dx$

or $=\ \int \left (\frac{4}{\Pi } \sin^{-1}\sqrt x - 1 \right )\ dx$

or $=\ \frac{4}{\Pi }\int \sin^{-1}\sqrt x - 1 \ dx\ -\ \int 1 \ dx\ +\ C$

or $=\ \frac{4}{\Pi }\int \sin^{-1}\sqrt x \ dx\ -\ x +\ C$

Thus $I\ =\ \frac{4}{\Pi }I'\ -\ x +\ C$

Now we will solve I'.

$I'\ =\ \int \sin^{-1}\sqrt x \ dx$

Put x = t ² .

Differentiating the equation wrt x, we get

$dx\ =\ 2t\ dt$

Thus $\int \sin^{-1}\sqrt x \ dx\ =\ \int \sin^{-1} t\ 2t \ dt$

or $=\ 2 \int t\ \sin^{-1} t\ \ dt$

Using integration by parts, we get :

$=\ 2 \left [ \sin^{-1}t \int t\ dt\ -\ \int \left ( \left ( \frac{d}{dt} \sin^{-1} t \right ) \int t\ dt \right ) \right ]\ dt$

or $=\ t^2 \sin^{-1}t\ -\ \int \frac{t^2}{\sqrt{1-t^2}}\ dt\ +\ C'$

We know that

$\int \frac{- t^2}{\sqrt{1-t^2}}\ dt\ =\ \frac{t}{2}\sqrt{1-t^2}\ -\ \frac{1}{2}\ \sin^{-1}t$

Thus it becomes :

$I'\ =\ t^2\sin^{-1} t\ +\ \frac{t}{2}\sqrt{1-t^2}\ -\ \frac{1}{2}\ \sin^{-1}t$

So I come to be :-

$I\ =\ \frac{4}{\Pi }I'\ -\ x +\ C$

$I\ =\ \sin^{-1}\sqrt{x} \left [ \frac{2(2x-1)}{\Pi } \right ]\ +\ \frac{2\sqrt{x-x^2}}{\Pi }\ -\ x\ +\ C$

Question:20 Integrate the functions in Exercises 1 to 24.

$\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}}$

Answer:

Given,

$\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}}$ = I (let)

Let $x= cos^2\theta \implies dx = -2sin\theta cos\theta d\theta$

$And \sqrt x= cos\theta \implies \theta = \cos^{-1}\sqrt x$

using the above substitution we can write the integral as

$\\ I = \int \sqrt{\frac{1-\sqrt {cos^2\theta}}{1 +\sqrt {cos^2\theta}}}(-2\sin\theta\cos\theta)d\theta \\ = -\int \sqrt{\frac{1-cos\theta}{1 +cos\theta}}(2\sin\theta\cos\theta)d\theta$

$\\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2\sin\theta\cos\theta)d\theta \\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2. 2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}\cos\theta)d\theta \\ = -4\int \sin^2\frac{\theta}{2}\cos\theta d\theta$

$\\ = -4\int \sin^2\frac{\theta}{2}(2cos^2\frac{\theta}{2} -1) d\theta$

1554447146338657

1554447147081166

1554447147826729

1554447148562798

1554447149316945

1554447150058843

1554447150798562

1554447151538561

1554447152324930

1554447153875620

Question:21 Integrate the functions in Exercises 1 to 24.

$\frac{2 + \sin 2x}{1 + \cos 2x}e^x$

Answer:

Given to evaluate

$\frac{2 + \sin 2x}{1 + \cos 2x}e^x$

1554447156109418

1554447156855423

1554447157709943

1554447158484136

1554447159245239

now the integral becomes

1554447160001344

Let tan x = f(x)

$\implies f'(x) = sec^2x dx$

1554447160771541

1554447161556589

1554447162309720

Question:22 Integrate the functions in Exercises 1 to 24.

$\frac{x^2 + x + 1}{(x+1)^2 (x+2)}$

Answer:

Given,

$\frac{x^2 + x + 1}{(x+1)^2 (x+2)}$

using partial fraction we can simplify the integral as

Let $\frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}$

$\\ \implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A(x+1)(x+2) + B(x+2) + C(x+1)^2}{(x+1)^2 (x+2)} \\ \implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A(x^2 + 3x+2) + B(x+2) + C(x^2 + 2x+1)}{(x+1)^2 (x+2)}$

$\\ \implies x^2 + x + 1 = A(x^2 + 3x+2) + B(x+2) + C(x^2 + 2x+1) \\ = (A+C)x^2 + (3A+B+2C)x + (2A+2B+C)$

Equating the coefficients of x, x ² and constant value, we get:

A + C = 1

3A + B + 2C = 1

2A+2B+C =1

Solving these:

A= -2, B=1 and C=3

$\implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}$

$\\ \implies \int \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \int\frac{-2}{x+1}dx+\int\frac{1}{(x+1)^2}dx+\int\frac{3}{x+2}dx \\ = -2\log(x+1) - \frac{1}{(x+1)} + 3\log (x+2) + C$

Question:23 Integrate the functions in Exercises 1 to 24.

$\tan^{-1}\sqrt{\frac{1-x}{1+x}}$

Answer:

We have

$I\ =\ \int \tan^{-1}\sqrt{\frac{1-x}{1+x}}$

Let us assume that : $x\ =\ \cos 2\Theta$

Differentiating wrt x,

$dx\ =\ -2 \sin 2\Theta\ d\Theta$

Substituting this in the original equation, we get

$\int \tan^{-1}\sqrt{\frac{1-x}{1+x}}\ =\ \int \tan^{-1}\sqrt{\frac{1-cos2\Theta }{1+cos2\Theta }}\times -2\sin 2\Theta \ d\Theta$

or $=\ -2\int \tan^{-1} (\frac{sin\Theta }{cos\Theta })\times \sin 2\Theta \ d\Theta$

or $=\ -2\int \Theta \sin 2\Theta \ d\Theta$

Using integration by parts , we get

$=\ -2\left ( \Theta \int \sin 2\Theta \ d\Theta\ - \int \frac{d\Theta }{d\Theta } \int \sin 2\Theta \ d\Theta\ \right )$

or $=\ -2\left ( \Theta \left ( \frac{-\cos 2\Theta }{2} \right ) - \int 1.\frac{-\cos 2\Theta }{2} \ d\Theta\ \right )$

or $=\ -2\left ( \frac{-\Theta \cos 2\Theta }{2}+ \frac{\sin 2\Theta }{4} \right )$

Putting all the assumed values back in the expression,

$=\ -2\left ( -\frac{1}{2}\left ( \frac{1}{2} \cos^{-1} x \right )+ \frac{\sqrt{1-x^2} }{4} \right )$

or $=\ \frac{1}{2}\left ( x \cos^{-1} x\ -\ \sqrt{1-x^2} \right )\ +\ C$

Question:24 Integrate the functions in Exercises 1 to 24.

$\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}$

Answer:

$\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}$

Here let's first reduce the log function.

$=\frac{\sqrt{x^2+1}}{x^4}\left [ \log (x^2+1)-\log x^2 \right ]dx$

$=\frac{\sqrt{x^2\left ( 1+\frac{1}{x^2} \right )}}{x^4}\left [ \log\frac{ (x^2+1)}{x^2} \right ]dx$

$=\int\frac{\sqrt{\left ( 1+\frac{1}{x^2} \right )}}{x^3}\left [ \log\left ( 1+\frac{1}{x^2} \right ) \right ]dx$

Now, let

$t=1+\frac{1}{x^2}$

$dt=\frac{-2}{x^3}dx$

So our function in terms if new variable t is :

$I=\frac{-1}{2}\int \left [\log t \right ]\cdot t^{\frac{1}{2}}dt$

now let's solve this By using integration by parts

$I=\frac{-1}{2}\int \left [(\log t)\frac{t^\frac{3}{2}}{\frac{3}{2}} -\int \frac{1}{t}\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}dt\right ]$

$I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\int t^{\frac{1}{2}}dt$

$I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}$

$I=\frac{2}{9}t^{\frac{3}{2}}-\frac{1}{3}t^{\frac{3}{2}}logt+c$

$I=\frac{1}{3}t^{\frac{3}{2}}\left [ \frac{2}{3}-\log t \right ]+c$

$I=\frac{1}{3}\left ( 1+\frac{1}{x^2} \right )^{\frac{3}{2}}\left [ \frac{2}{3}-\log \left ( 1+\frac{1}{x^2} \right ) \right ]+c$

Question:25 Evaluate the definite integrals in Exercises 25 to 33.

$\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx$

Answer:

$\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx$

Since, we have $e^x$ multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,

$\int e^x(f(x)+f'(x))dx=e^xf(x)$

So,

$\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx$

$=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-2\sin \frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}} \right )dx$

$=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2sin^2\frac{x}{2}} -\frac{2\sin \frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}}\right )dx$

$=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2}cosec^2\frac{x}{2}-cot\frac{x}{2}\right )dx$

$=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx$

Here let's use the property

$\int e^x(f(x)+f'(x))dx=e^xf(x)$

so,

$=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx$

$=\left [ -e^xcot\frac{x}{2} \right ]_\frac{\pi}{2}^\pi$

$=\left [ -e^\pi cot\frac{\pi}{2} \right ]-\left [ -e^{\frac{\pi}{2}} cot\frac{\pi}{4} \right ]$

$=e^{\frac{\pi}{2}}$

Question:26 Evaluate the definite integrals in Exercises 25 to 33.

$\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}$

Answer:

$\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}$

First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)

Let' divide both numerator and denominator by $cos^4x$

$=\int_0^\frac{\pi}{4}\frac{\frac{\sin x\cos x}{cosxcosxsos^2x} }{1+\frac{\sin^4 x}{\cos^4x}}$

$=\int_0^\frac{\pi}{4}\frac{tanxsec^2x}{1+tan^4x}$

Now lets change the variable

$\\t=tan^2x \\dt=2tanxsec^2xdx$

the limits will also change since the variable is changing

$when\:x=0,t=tan^20=0$

$when\:x=\frac{\pi}{4},t=tan^2\frac{\pi}{4}=1$

So, the integration becomes:

$I=\frac{1}{2}\int_{0}^{1}\frac{dt}{1+t^2}$

$I=\frac{1}{2}\left [ tan^{-1}t \right ]_0^1$

$I=\frac{1}{2}\left [ tan^{-1}1 \right ]-\frac{1}{2}\left [ tan^{-1}0\right ]$

$I=\frac{1}{2}\left [ \frac{\pi}{4} \right ]-0$

$I=\frac{\pi}{8}$

Question:27 Evaluate the definite integrals in Exercises 25 to 33.

$\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}$

Answer:

Lets first simplify the function.

$\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4(1-\cos^2 x)}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{4-3\cos^2 x}$

$\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x-4\:\: }{4-3\cos^2 x }dx=\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x\:\: }{4-3\cos^2 x }dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx$

$\\=\frac{-1}{3}\int_0^\frac{\pi}{2}1dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx \\ \\ \\=\frac{-1}{3} \left [ x \right ]_0^{\frac{\pi}{2}}-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4\:\: }{4-3\cos^2 x }dx$

As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,

$=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4sec^2x-3 }dx$

AS we can write square of sec in term of tan,

$=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4(1+tan^2x)-3 }dx$

$=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx$

Now let's calculate the integral of the second function, (we already have calculated the first function)

$=-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx$

let

$\\t=2tanx, \\dt=2sec^2xdx$

here we are changing the variable so we have to calculate the limits of the new variable

when x = 0, t = 2tanx = 2tan(0)=0

when $x=\pi/2,t=2tan{\pi/2}=\infty$

our function in terms of t is

$=-\frac{2}{3}\int_0^\infty\frac{1 }{1+t^2 }dt$

$=\left [ tan^{-1} t\right ]_0^\infty=\left [ tan^{-1} \infty-tan^{-1} 0\right ]$

$=\frac{\pi}{2}$

Hence our total solution of the function is

$\\=-\frac{\pi}{6}+\frac{2}{3}*\frac{\pi}{2}\\=\frac{\pi}{6}$

Question:28 Evaluate the definite integrals in Exercises 25 to 33.

$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}$

Answer:

$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}$

Here first let convert sin2x as the angle of x ( sinx, and cosx)

$\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{2sinxcosx}}$

Now let's remove the square root form function by making a perfect square inside the square root

$\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{-(-1+1-2sinxcosx)}}$

$\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{(1-(sin^2x+cos^2x-2sinxcosx)}}$

$\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{(1-(sinx-cosx)^2}}$

Now let

, $\\t=sinx-cosx \\dt=(cosx+sinx)dx$

since we are changing the variable, limit of integration will change

$\\when\: x=\pi/6, t=sin\pi/6-cos\pi/6=(1-\sqrt{3})/2 \\ when x= \pi/3,t=sin\pi/3-cos/pi/3=(\sqrt{3}-1)/2$

our function in terms of t :

$\\=\int_\frac{1-\sqrt{3}}{2}^\frac{\sqrt{3}-1}{2} \frac{1}{\sqrt{(1-t^2)}}dt$

$\\=\left [ sin^{-1}t \right ]_\frac{1-\sqrt{3}}{2}^\frac{\sqrt{3}-1}{2} \\ \\ \\=2sin^{-1}\left (\frac{\sqrt{3}-1}{2} \right )$

Question:29 Evaluate the definite integrals in Exercises 25 to 33.

$\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}$

Answer:

$\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}$

First, let's get rid of the square roots from the denominator,

$\\=\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}*\frac{\sqrt{1+x} +\sqrt x}{\sqrt{1+x} +\sqrt x}$

$\\=\int_0^1\frac{\sqrt{1+x}+\sqrt{x}}{{1+x} -x}dx$

$\\=\int_0^1({\sqrt{1+x}+\sqrt{x}})dx$

$\\=\int_0^1({\sqrt{1+x})dx+\int_0^1({\sqrt{x}})dx$

$\\=\int_0^1(1+x)^\frac{1}{2}dx+\int_0^1x^\frac{1}{2}dx$

$\\=\left [ \frac{2}{3}(1+x)^{\frac{3}{2}} \right ]_0^1+\left [ \frac{2}{3}(x)^{\frac{3}{2}} \right ]_0^1$

$\\=\left [ \frac{2}{3}(1+1)^{\frac{3}{2}} \right ]-\left [ \frac{2}{3} \right ]+\left [ \frac{2}{3}(1)^{\frac{3}{2}} \right ]-\left [ 0 \right ]$

$\\=\frac{4\sqrt{2}}{3}$

Question:30 Evaluate the definite integrals in Exercises 25 to 33.

$\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx$

Answer:

$\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx$

First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt

So,

Now since we are changing the variable, the new limit of the integration will be,

when x = 0, t = cos0-sin0=1-0=1

when $x=\pi/4$ $t=\cos\pi/4-\sin\pi/4=0$

Now,

$(\cos x-\sin x)^2=t^2$

$\cos ^2x+\sin^2 x-2\cos x \sin x =t^2$

$1-\sin 2x =t^2$

$\sin 2x =1-t^2$

Hence our function in terms of t becomes,

$\int_{-1}^{0}\frac{dt}{9+16(1-t^2)}=\int_{-1}^{0}\frac{dt}{9+16-16t^2}=\int_{-1}^{0}\frac{dt}{25-16t^2}=\int_{-1}^{0}\frac{dt}{5^2-(4t)^2)}$

$= \frac{1}{4}\left [\frac{1}{2(5)}\log \frac{5+4t}{5-4t} \right ]_{-1}^0$

$= \frac{1}{40}\left[ \log (1)-\log (\frac{1}{9})\right ]$

$=\frac{\log 9}{40}$

Question:31 Evaluate the definite integrals in Exercises 25 to 33.

$\int_0^\frac{\pi}{2}\sin 2x\tan^{-1}(\sin x)dx$

Answer:

Let I =

$\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$

$=\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$

Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so

$\\t=sinx \\dt=cosxdx$

Now the important step here is to change the limit of the integration as we are changing the variable.so,

$\\when\:x=0,t=sin0=0 \\when\:x=\frac{\pi}{2},t=sin\frac{\pi}{2}=1$

So our function becomes,

$I=2\int_{0}^{1}(tan^{-1}t)tdt$

Now, let's integrate this by using integration by parts method,

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\int\frac{1}{1+t^2}\cdot\frac{t^2}{2}dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{t^2}{1+t^2}\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{(1+t^2)-1}{1+t^2}\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\left ( 1-\frac{1}{1+t^2} \right )\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t-tan^{-1}t) \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t)+\frac{1}{2}tan^{-1}t) \right ]_0^1$

$I=2\left [ \frac{1}{2} \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1$

$I=\left [ \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1$

$I=\left [ \left (tan^{-1}(1)\cdot(1^2+1)-1 \right )\right ]-\left [ \left (tan^{-1}(0)\cdot(0^2+1)-0 \right )\right ]$ $I=2tan^{-1}1-1=2\times \frac{\pi}{4}-1$

$I=\frac{\pi}{2}-1$

Question:32 Evaluate the definite integrals in Exercises 25 to 33.

$\int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx$

Answer:

Let I = $\int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx$ -(i)

Replacing x with ( $\pi$ -x),

$\\ I = \int_\pi^0\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x) + \tan (\pi -x)} (-dx) \\ = -\int_\pi^0\frac{(\pi -x)(-)\tan x}{-\sec x - \tan x} dx$

$\\ \implies I = \int^\pi_0\frac{(\pi -x)\tan x}{\sec x + \tan x} dx$ - (ii)

Adding (i) and (ii)

$I + I = \int^\pi_0\left(\frac{x\tan x}{\sec x + \tan x} + \frac{(\pi -x)\tan x}{\sec x + \tan x} \right) dx$

$\implies 2I = \int^\pi_0\frac{\pi\tan x}{\sec x + \tan x} dx$

$\\ \implies 2I = \int^\pi_0\frac{\pi \frac{sin x}{cos x} }{\frac{1}{cos x} + \frac{sin x}{cos x}} dx \\ \implies 2I =\pi \int^\pi_0\frac{ sin x }{1+sin x} dx \\ \implies 2I =\pi \int^\pi_0\frac{ (1 +sin x ) -1}{1+sin x} dx \\ \implies 2I =\pi \int^\pi_0\left [1- \frac{1}{1+sin x} \right ]dx$

$\\ \implies 2I =\pi \int^\pi_0\left [1- \frac{1}{1+sin x} \right ]dx \\ \implies 2I =\pi \int^\pi_01 dx - \pi \int^\pi_0\frac{1}{1+sin x}.\frac{(1-sin x)}{(1 - sin x)}dx \\ \implies 2I =\pi\int^\pi_01 dx - \pi \int^\pi_0[\sec^2 x - \sec x \tan x]dx \\ \implies 2I =\pi[x]^\pi_0 - \pi[\sec x - \tan x]^\pi_0$

$\\ \implies 2I =\pi[\pi - 0] - \pi[tan \pi - sec \pi- tan \pi + sec 0] \\ \implies 2I =\pi[\pi -2] \\ \implies I =\frac{\pi}{2}[\pi -2]$

Question:33 Evaluate the definite integrals in Exercises 25 to 33.

$\int_1^4[|x-1| + |x-2| + |x-3|]dx$

Answer:

Given integral $\int_1^4[|x-1| + |x-2| + |x-3|]dx$

So, we split it in according to intervals they are positive or negative.

$= \int_{1}^4 |x-1| dx + \int_{1}^4 |x-2| dx + \int^4_{1} |x-3| dx$

$= I_{1}+I_{2}+I_{3}$

Now,

$I_{1} = \int^4_{1}|x-1| dx = \int^4_{1} (x-1)dx$

$\because$ as $(x-1)$ is positive in the given x -range $[1,4]$

$=\left [ \frac{x^2}{2}-x\right ]^4_{1} = \left [ \frac{4^2}{2}-4 \right ] - \left [ \frac{1^2}{2}-1 \right ]$

$= \left [ 8-4 \right ] - [-\frac{1}{2}] = 4+\frac{1}{2} = \frac{9}{2}$

Therefore, $I_{1} = \frac{9}{2}$

$I_{2} = \int^4_{1}|x-2| dx = \int^2_{1} (2-x)dx +\int^4_{2} (x-2)dx$

$\because$ as $(x-2)\geq 0$ is in the given x -range $[2,4]$ and $\leq 0$ in the range $[1,2]$

$=\left [ 2x - \frac{x^2}{2}\right ] ^2_{1} + \left [ \frac{x^2}{2} -2x\right ] ^4_{2}$

$= \left \{ \left [ 2(2)-\frac{2^2}{2} \right ] - \left [ 2(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-2(4) \right ] - \left [ \frac{2^2}{2}-2(2) \right ] \right \}$

$= [4-2-2+\frac{1}{2}] +[8-8-2+4]$

$= \frac{1}{2}+2 =\frac{5}{2}$

Therefore, $I_{2} = \frac{5}{2}$

$I_{3} = \int^4_{1}|x-3| dx = \int^3_{1} (3-x)dx +\int^4_{3} (x-3)dx$

$\because$ as $(x-3)\geq 0$ is in the given x -range $[3,4]$ and $\leq 0$ in the range $[1,3]$

$=\left [ 3x - \frac{x^2}{2}\right ] ^3_{1} + \left [ \frac{x^2}{2} -3x\right ] ^4_{3}$

$= \left \{ \left [ 3(3)-\frac{3^2}{2} \right ] - \left [ 3(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-3(4) \right ] - \left [ \frac{3^2}{2}-3(3) \right ] \right \}$

$= [9-\frac{9}{2}-3+\frac{1}{2}]+[8-12-\frac{9}{2}+9]$

$= [6-4]+\frac{1}{2} =\frac{5}{2}$

Therefore, $I_{3} = \frac{5}{2}$

So, We have the sum $= I_{1}+I_{2}+I_{3}$

$I = \frac{9}{2}+\frac{5}{2}+\frac{5}{2} = \frac{19}{2}$

Question:34 Prove the following (Exercises 34 to 39)

. $\int_1^3\frac{dx}{x^2(x+1)} = \frac{2}{3}+ \log \frac{2}{3}$

Answer:

L.H.S = $\int_1^3\frac{dx}{x^2(x+1)}$

We can write the numerator as [(x+1) -x]

$\therefore \int_1^3\frac{dx}{x^2(x+1)} = \int_1^3\frac{(x+1)-x}{x^2(x+1)}dx$

$\\ = \int_1^3\left [ \frac{1}{x^2} - \frac{1}{x(x+1)} \right ]dx \\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{(x+1)-x}{x(x+1)}dx$

$\\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\left [ \frac{1}{x} - \frac{1}{(x+1)} \right ]dx \\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{1}{x}dx + \int_1^3\frac{1}{(x+1)}dx \\ = \left [ -\frac{1}{x} \right ]^3_1 - \left [ \log x \right ]^3_1 +\left [ \log(x+1) \right ]^3_1$

$\\ = \left [ -\frac{1}{3} +1 \right ] - \left [ \log 3 - \log 1 \right ] +\left [\log 4 - \log 2 \right ] \\ = \frac{2}{3} + \log \left ( \frac{4}{3.2}\right ) \\$

$= \log \left(\frac{2}{3} \right ) +\frac{2}{3}$ = RHS

Hence proved.

Question:35 Prove the following (Exercises 34 to 39)

$\int_0^1 xe^xdx =1$

Answer:

$Let\ I=\int xe^{x}dx$

Integrating I by parts

$\\I=x\int e^{x}dx-\int ( (\frac{\mathrm{d} (x)}{\mathrm{d} x})\int e^{x}dx)dx\\ I=xe^{x}-\int e^{x}dx\\ I=xe^{x}-e^{x}+c$

Applying Limits from 0 to 1

$\\\int_{0}^{1}xe^{x}dx=[xe^{x}-e^{x}+c]_{0}^{1}\\ I=[e-e+c]-[0-1+c]\\ I=1$

Hence proved I = 1

Question:36 Prove the following (Exercises 34 to 39)

$\int_{-1}^1x^{17}\cos^4 x dx=0$

Answer:

$Let \ x^{17}cos^{4}x=g(x)$

$g(-x)= (-x)^{17}cos^{4}(-x)=-x^{17}cos^{4}x=-g(x)$

The Integrand g(x) therefore is an odd function and therefore

$\int_{-1}^{1}g(x)dx=0$

Question:37 Prove the following (Exercises 34 to 39)

$\int_0^\frac{\pi}{2}\sin^3 x dx =\frac{2}{3}$

Answer:

$\\Let\ I= \int_{0}^{\frac{\pi }{2}}sin^{3}xdx\\ I=\int_{0}^{\frac{\pi }{2}}sinx(1-sin^{2}x)dx\\ I=\int_{0}^{\frac{\pi }{2}}sinxdx-\int_{0}^{\frac{\pi }{2}}cos^{2}xsinxdx\\ I=I_{1}-I_{2}$

$\\I_{1}=[-cosx]_{0}^{\frac{\pi }{2}}\\ I_{1}=-0-(-1)=1$

For I ₂ let cosx=t, -sinxdx=dt

The limits change to 0 and 1

$\\I_{2}=-\int_{1}^{0}t^{2}dt\\ I_{2}=-[\frac{t^{3}}{3}]{_{1}}^{0}\\ I_{2}=0-(-\frac{1}{3})\\ I_{2}=\frac{1}{3}$

I ₁ -I ₂ =2/3

Hence proved.

Question:38 Prove the following (Exercises 34 to 39)

$\int_0^\frac{\pi}{4}2\tan^3 x dx = 1 - \log 2$

Answer:

The integral is written as

$\\Let\ I=\int 2tan^{3}xdx\\ I=\int 2tan^{2}x\cdot tanxdx\\ I=\int 2(sec^{2}x-1)tanxdx\\ I=2\int tanxsec^{2}xdx-2\int tanxdx\\ I=2\int tdt-2log(cosx)+c\ \ \ \ \ \ \ (t=tanx) \\I=t^{2}-2log(cosx)+c\\ I=tan^{2}x-2log(cosx)+c$

$[I]_{0}^{\frac{\pi }{4}}=[tan^{2}x-2log(cosx)]_{0}^{\frac{\pi }{4}}\\$

$[I]_{0}^{\frac{\pi }{4}}=(1-2log\sqrt{2})-(0-2log1)$

$[I]_{0}^{\frac{\pi }{4}}=1-log2$

Hence Proved

Question:39 Prove the following (Exercises 34 to 39)\

$\int_0^1\sin^{-1}xdx = \frac{\pi}{2}-1$

Answer:

$Let \ I=\int sin^{-}xdx$

Integrating by parts we get

$\\ I= sin^{-}x\int 1\cdot dx-\int (\frac{\mathrm{d} (sin^{-}x)}{\mathrm{d} x}\int 1\cdot dx)\\ I=xsin^{-}x+c-\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx\\ I=I_{1}-I_{2}$

For I ₂ take 1-x ² = t ² , -xdx=tdt

$\\I_{2}=\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx\\ I_{2}=-\int\frac{1}{t}tdt \\ I_{2}=-t+c\\ I_{2}=-\sqrt{1-x^{2}}+c$

$[I]_{0}^{1}=[I_{1}-I_{2}]_{0}^{1}\\$

$\\=[xsin^{-}x-(-\sqrt{1-x^{2}})]_{0}^{1}\\ =[xsin^{-}x+\sqrt{1-x^{2}}]_{0}^{1}\\ =[1\cdot \frac{\pi }{2}+0]-[0+1]\\ =\frac{\pi }{2}-1$

Hence Proved

Question:40 Evaluate $\int_0^1e^{2-3x}dx$ as a limit of a sum.

Answer:

As we know

$\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)........+f(a+(n-1)h)]$

where b-a=hn

In the given problem b=1, a=0 and $f(x)=e^{2-3x}$
$\\\int_{0}^{1}e^{2-3x}dx=(1-0)\lim_{n\rightarrow \infty }\frac{1}{n}(e^{2}+e^{2-3h}+e^{2-3(2h)}.....+e^{2-3(n-1)h})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(1+e^{-3h}+e^{-6h}....+e^{-3(n-1)h})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-(e^{-3h})^{n}}{1-e^{-3h}})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-e^{-\frac{3}{n}\times n}}{1-e^{-\frac{3}{n}}})\\$

$\\=e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-e^{-3}}{1-e^{-\frac{3}{n}}})\\ =\frac{e^{2}(1-e^{-3})}{3}\lim_{n\rightarrow \infty }\frac{-\frac{3}{n}}{e^{-\frac{3}{n}}-1}\\ =\frac{e^{2}(1-e^{-3})}{3}$

$=\frac{e^{2}-e^{-1}}{3}$

Question:41 Choose the correct answers in Exercises 41 to 44.

. $\int\frac{dx}{e^x + e^{-x}}$ is equal to

(A) $\tan^{-1}(e^x) + c$

(B) $\tan^{-1}(e^{-x}) + c$

(D) $\log (e^x + e^{-x}) + C$

Answer:

$\int\frac{dx}{e^x + e^{-x}}$

the above integral can be re arranged as

$\\=\int \frac{e^{x}}{e^{2x}+1}dx\\$

let e ^x =t, e ^x dx=dt

$\int\frac{dx}{e^x + e^{-x}}$

$\\=\int \frac{1}{t^{2}+1}dt\\ =tan^{-1}t+c\\ =tan^{-1}(e^{x})+c$

(A) is correct

Question:42âââââââ Choose the correct answers in Exercises 41 to 44.

. $\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx$ is equal to

(A) $\frac{-1}{\sin x + \cos x} + C$

(B) $\log |{\sin x + \cos x} |+ C$

(D) $\frac{1}{(\sin x + \cos x)^2} + C$

Answer:

$\\\frac{\cos 2x}{(\sin x + \cos x)^2}\\ =\frac{cos^{2}x-sin^{2}x}{(\sin x + \cos x)^2}\\ =\frac{(\sin x + \cos x)(\cos x-\sin x)}{(\sin x + \cos x)^2} \\=\frac{(\cos x-\sin x)}{(\sin x + \cos x)}$ cos2x=cos ² x-sin ² x

let sinx+cosx=t,(cosx-sinx)dx=dt

hence the given integral can be written as

$\\\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx\\ =\int \frac{dt}{t}\\ =log|t|+c \\=log|cosx+sinx|+c$

B is correct

Question:43âââââââ Choose the correct answers in Exercises 41 to 44.

If $f(a+b-x) = f(x)$ , then $\int_a^bxf(x)dx$ is equal to

(A) $\frac{a+b}{2}\int^b_af(b-x)dx$

(B) $\frac{a+b}{2}\int^b_af(b+x)dx$

(D) $\frac{a+b}{2}\int^b_af(x)dx$

Answer:

$Let\ \int_a^bxf(x)dx=I$

As we know $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

Using the above property we can write the integral as

$\\I=\int_{a}^{b}(a+b-x)f(a+b-x)dx\\ I=\int_{a}^{b}(a+b-x)f(x)dx\\ I=(a+b)\int_{a}^{b}f(x)dx-\int_{a}^{b}xf(x)dx\\ I=(a+b)\int_{a}^{b}f(x)dx-I\\ 2I=(a+b)\int_{a}^{b}f(x)dx\\ I=\frac{a+b}{2}\int_{a}^{b}f(x)dx$

Answer (D) is correct

Question:44 Choose the correct answers in Exercises 41 to 44.

The value of $\int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx$ is

(A) 1

(B) 0

(D) $\frac{\pi}{4}$

Answer:

$\\Let\ I=\int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx\\$

$\\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )\\ =tan^{-1}\left ( \frac{x-(1-x)}{1+x(1-x)} \right )\\ =tan^{-1}x-tan^{-1}(1-x)$ as $tan^{-1}\left ( \frac{a-b}{1+ab} \right )=tan^{-1}a-tan^{-1}b$

Now the integral can be written as

$\\I=\int_{0}^{1} \left ( tan^{-1}x-tan^{-1}(1-x) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}(1-(1-x)) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}x\right )dx\\ I=-I\\ 2I=0\\ I=0$

(B) is correct.

If you are looking for integrals class 12 NCERT solutions of exercises then they are listed below.

About NCERT solutions for class 12 maths chapter 7 integrals

The word integration literally means summation. When you have to find the sum of finite numbers you can do by simply adding these numbers. But when you are finding the sum of a certain number of elements as the number of elements tends to infinity and at the same time each term becomes infinitesimally small, you can use a prosses to find its limit called integration.

Integrals has 13 % weightage in 12 board final examinations. Next chapter "applications of integrals" is also dependent on this chapter. So you should try to solve every problem of this chapter on your own. If you are not able to do, you can take the help of these NCERT solutions for class 12 maths chapter 7 integrals. In this chapter, there are 11 exercises with 227 questions and also 44 questions are there in miscellaneous exercise. Here, the NCERT solutions for class 12 maths chapter 7 integrals are solved and explained in detail to develop a grip on the topic. Here, you will learn two types of integrals: Definite integral and Indefinite integral and also learn their properties and formulas.

	Definite Integral	Indefinite Integral
Definition	A definite Integral has upper and lower limits if 'a' and 'b' are the limits or boundaries. The definite integral of f(x) is a number, not function .	An integral without upper limit and lower limit. It is also an antiderivative. The indefinite integral of f(x) is a function not number.
Expression	$\int_{a}^{b}f(x)dx$	$F(x)=\int f(x)dx$

NCERT Solutions for Class 12 Maths Chapter 7 Integrals - Topics

7.1 Introduction

7.2 Integration as an Inverse Process of Differentiation

7.2.1 Geometrical interpretation of indefinite integral

7.2.2 Some properties of indefinite integral

7.2.3 Comparison between differentiation and integration

7.3 Methods of Integration

7.3.1 Integration by substitution

7.3.2 Integration using trigonometric identities

7.4 Integrals of Some Particular Functions

7.5 Integration by Partial Fractions

7.6 Integration by Parts

7.7 Definite Integral

7.7.1 Definite integral as the limit of a sum

7.8 Fundamental Theorem of Calculus

7.8.1 Area function

7.8.2 First fundamental theorem of integral calculus

7.8.3 Second fundamental theorem of integral calculus

7.9 Evaluation of Definite Integrals by Substitution

7.10 Some Properties of Definite Integral

NCERT solutions for class 12 maths - Chapter wise

Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

Key Features of NCERT Solutions for Class 12 Maths Chapter 7 Integrals

The NCERT chapter 7 class 12 maths offers several key features to aid students in their understanding and mastery of this topic. Some of these features include

Detailed explanation: The chapter 7 class 12th maths solutions provide a comprehensive and in-depth explanation of the concepts of integrals, making it easier for students to grasp the subject.
Step-by-Step Solutions: The integration class 12 ncert solutions break down complex problems into simpler steps, making it easier for students to follow and understand the process.
Clear Diagrams: The integrals class 12 solutions make use of clear and informative diagrams to help students visualize the concepts, making it easier for them to comprehend the material.
Plenty of Practice Problems: The ch 7 maths class 12 include a large number of practice problems to help students strengthen their knowledge and skills.
Accurate answers: The class 12 ch 7 maths ncert solutions are checked and verified by experts to ensure accuracy, helping students avoid mistakes and improve their grades.

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NCERT solutions for class 12 - subject wise

How to use NCERT solutions for class 12 maths chapter 7 Integrals

You are expected to remember all formulas of differentiation, and then can start with basic integration
Try to relate differentiation formulas with integrations formulas because it will help you to remember all the integration formulas
When you are done with basic integration, you should learn different methods of integration for different types of functions. If you find difficulties in learning the method you should learn one method at a time to solve a particular type of problem
After learning the different methods of integration, you should be able to determine which method will be used to solve a particular type of problem
When you become good with indefinite integrals, move to the definite integrals and learn some properties to solve definite integrals. NCERT Class 12 Maths solutions chapter 7 integrals will help you for the same
This chapter requires a lot of practice. First, solve all the NCERT textbook questions, then, you can take the help of NCERT solutions for class 12 maths chapter 7 integrals.
If you have solved all NCERT questions, you can solve CBSE previous year papers also to get familiar with the type of questions which are asked in previous years.
NCERT solutions class 12 maths chapter 12 pdf download will be made available soon. Till then you can save the webpage and practice these solutions offline.

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. Which are the most difficult chapters of NCERT Class 12 Maths syllabus?

Students consider Integration which is integrals and applications of integration are the most difficult chapter in CBSE class 12 maths but with the regular practice of NCERT problems you will be able to have a strong grip on this chapter also. it's true that there is no substitute for hard work but the right strategy and quality study material are also essential to get command of this chapter, therefore, NCERT exercises are recommended for practice.

2. How are the NCERT solutions helpful in the board exam?

NCERT solutions are created by the expert team of careers360 who know how best to write answers in the board exam in order to get good marks. Integrating their techniques can benefit in obtaining meritorious marks. Sometimes students do not understand where s/he are making a mistake, NCERT solutions can help them to understand that. the practice of a lot of questions and their solution make you confident and help you in getting an in-depth understanding of concepts. Therefore NCERT solutions are very helpful for students.

3. What is the weightage of the chapter Integrals for CBSE board exam?

Integrals have 13 % weightage in CBSE class 12th board final examination. having 13% weightage, Integral become very students for CBSE aspirant but it demand lot of practice. Interested students can refer to integrals class 12 solutions.

4. Is the NCERT Solutions for class 12 maths integration challenging to comprehend?

The NCERT Solutions for maths ch 7 maths class 12 is not complicated to understand. It is a fascinating topic in Class 12 that is also relevant at higher education levels. A solid understanding of integral formulas will enable students to solve integration problems effectively. A thorough knowledge of derivatives is crucial for comprehending the concepts of integral calculus with ease. For ease, students can study integrals class 12 ncert solutions pdf both online and offline.

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Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

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Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

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Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

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Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals

NCERT Integrals Class 12 Questions And Answers

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Class 12 maths chapter 7 NCERT Solutions - Important Formulae

NCERT Integrals Class 12 Questions And Answers (Intext Questions and Exercise)

Answer:

Question:2 Find an anti derivative (or integral) of the following functions by the method of inspection.

Answer:

Question:3 Find an anti derivative (or integral) of the following functions by the method of inspection.

Answer:

Question:4 Find an anti derivative (or integral) of the following functions by the method of inspection.

Answer:

Question:5 Find an anti derivative (or integral) of the following functions by the method of inspection.

Answer:

Question:6 Find the following integrals

Question:2 Find an anti derivative (or integral) of the following functions by the method of inspection. $\cos 3x$

Question:3 Find an anti derivative (or integral) of the following functions by the method of inspection. $e ^{2x}$

Question:4 Find an anti derivative (or integral) of the following functions by the method of inspection. $( ax + b )^2$

Question:5 Find an anti derivative (or integral) of the following functions by the method of inspection. $\sin 2x - 4 e ^{3x}$

Question:7 Find the following integrals $\int x ^2 ( 1- \frac{1}{x^2})dx$

Question:8 Find the following integrals $\int ( ax ^2 + bx + c ) dx$

Question:9 Find the following integrals intergration of $\int \left ( 2x^2 + e ^x \right ) dx$

Question:10 Find the following integrals $\int \left ( \sqrt x - \frac{1}{\sqrt x } \right ) ^2 dx$

Question:11 Find the following integrals intergration of $\int \frac{x^3 + 5x^2 - 4}{x^2} dx$

Question:12 Find the following integrals $\int \frac{x^3+ 3x +4 }{\sqrt x } dx$

Question:13 Find the following integrals intergration of $\int \frac{x^3 - x^2 + x -1 }{x-1 } dx$

Question:14 Find the following integrals $\int (1-x) \sqrt x dx$

Question:15 Find the following integrals $\int \sqrt x ( 3x^2 + 2x +3 )dx$

Question:16 Find the following integrals $\int ( 2x - 3 \cos x + e ^x ) dx$

Question:17 Find the following integrals $\int ( 2 x ^2 - 3 \sin x + 5 \sqrt x ) dx$

Question:18 Find the following integrals $\int \sec x ( \sec x + \tan x ) dx$

Question:19 Find the following integrals intergration of $\int \frac{sec ^2 x }{cosec ^2 x } dx$

Question:20 Find the following integrals $\int \frac{2- 3 \sin x }{\cos ^ 2 x } dx$

Question:21 Choose the correct answer
The anti derivative of $\left ( \sqrt x + 1/ \sqrt x \right )$ equals

Question:1 Integrate the functions $\frac{2x}{1+ x ^2}$

Question:2 Integrate the functions $\frac{( \log x )^2}{x}$

Question:3 Integrate the functions $\frac{1}{x+ x \log x }$

Question:4 Integrate the functions $\sin x \sin ( \cos x )$

Question:5 Integrate the functions $\sin ( ax + b ) \cos ( ax + b )$

Question:6 Integrate the functions $\sqrt { ax + b }$

Question:7 Integrate the functions $x \sqrt { x +2 }$

Question:8 Integrate the functions $x \sqrt { 1+ 2 x^2 }$

Question:9 Integrate the functions $( 4x +2 ) \sqrt { x ^ 2 + x + 1 }$

Question:10 Integrate the functions $\frac{1}{x - \sqrt x }$

Question:11 Integrate the functions $\frac{x }{ \sqrt{ x +4} }$ , x > 0

Question:12 Integrate the functions $( x ^3 - 1 ) ^{1/3} x ^ 5$

Question:13 Integrate the functions $\frac{x ^2 }{(2+3x^3)^3}$

Question:14 Integrate the functions $\frac{1}{x (\log x )^m} , x > 0 , m \neq 1$

Question:15 Integrate the functions $\frac{x}{9- 4 x ^2 }$

Question:16 Integrate the functions $e ^{ 2 x +3 }$

Question:17 Integrate the functions $\frac{x }{e^{x^{2}}}$

Question:18 Integrate the functions $\frac{e ^{\tan ^{-1}x}}{1+ x^2 }$