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  • IF  such that <img alt="n\epsilon N" src="https://lea

IF a_nI_{n+2}+b_nI_n=C_n such that n\epsilon N and I_n=\int_{0}^{1}x^n\tan^{-1}x dx, then 

Option: 1

a1,a2,a3.... are in A.P.


Option: 2

b1,b2,b3.... are in G.P.


Option: 3

c1,c2,c3.... are in H.P.


Option: 4

a1,a2,a3.... are in H.P.


Answers (1)

best_answer

 

 

Definite Integration -

Let f be a function of x defined on the closed interval [a, b] and F be another function such that \frac{d}{dx}(F(x))=f(x)   for all x in the domain of f, then 

\int_{a}^{b} f(x) d x=[F(x)+c]_{a}^{b}=F(b)-F(a)

is called the definite integral of the function f(x) over the interval [a, b],  where a is called the lower limit of the integral and b is called the upper limit of the integral. 

-

I_n=\int_{0}^{1}x^n\tan^{-1}x dx

I_{n+2}=\int_{0}^{1}x^{n+2}\tan^{-1}x dx

I_{n+2}=\int_{0}^{1}x^{n+1}\cdot x\tan^{-1}x dx

I_{n+2}=x^{n+1}\cdot \int_{0}^{1}x\tan^{-1}x dx-\int_{0}^{1}\left [ \frac{d}{dx}\left ( x^{n+1} \right )\cdot \int_{0}^{1}x\tan^{-1}x dx \right ]dxSolve for I=\int_{0}^{1}x\tan^{-1}x dx

I=\tan^{-1}x \int xdx-\int \left [\frac{d}{dx}(\tan^{-1}x)\cdot \int xdx \right ]dx

I= \frac{x^2}{2}\tan^{-1}x-\int \left [\frac{1}{1+x^2}\cdot \frac{x^2}{2} \right ]dx

I= \frac{x^2}{2}\tan^{-1}x-\frac{1}{2} \int \left [\frac{x^2+1-1}{1+x^2} \right ]dx

I= \frac{x^2}{2}\tan^{-1}x-\frac{1}{2} \int 1 dx+ \frac12\int \left [\frac{1}{1+x^2} \right ]dx

I= \frac{x^2}{2}\tan^{-1}x-\frac{x}{2} +\frac12 \tan^{-1}x

Hence we can rewrite I_{n+2} as,

I_{n+2}=x^{n+1}\cdot I-\int\left [ \frac{d}{dx}\left ( x^{n+1} \right )\cdot I \right ]dx

I_{n+2}=x^{n+1}\cdot \left [ \frac{x^2}{2}\tan^{-1}x-\frac{x}{2} +\frac12 \tan^{-1}x \right ]-\left (n+1 \right )\int\left[ x^{n}\cdot \left ( \frac{x^2}{2}\tan^{-1}x-\frac{x}{2} +\frac12 \tan^{-1}x \right ) \right ]dxI_{n+2}=x^{n+1}\cdot \left [ \frac{x^2}{2}\tan^{-1}x-\frac{x}{2} +\frac12 \tan^{-1}x \right ]- \frac{n+1}2 \int\left[ x^{n+2}\cdot \tan^{-1}x-x^{n+1} +x^n \tan^{-1}x \right ]dx\\I_{n+2}=x^{n+1}\cdot \left [ \frac{x^2}{2}\tan^{-1}x-\frac{x}{2} +\frac12 \tan^{-1}x \right ]- \frac{n+1}2 \left [ I_{n+2}\;\;-\int(x^{n+1})dx +I_n \right ]\\I_{n+2}=x^{n+1}\cdot \left [ \frac{x^2}{2}\tan^{-1}x-\frac{x}{2} +\frac12 \tan^{-1}x \right ]- \frac{n+1}2 \left [ I_{n+2}\;\;-\frac{x^{n+2}}{n+2} +I_n \right ]

Now apply limit

\\I_{n+2}=\left [ x^{n+1}\cdot \left ( \frac{x^2}{2}\tan^{-1}x-\frac{x}{2} +\frac12 \tan^{-1}x\right ) \right ]^{1}_{0}- \frac{n+1}2 \left [ I_{n+2}\;\;-\frac{x^{n+2}}{n+2} +I_n \right ]^{1}_{0}

\\I_{n+2}=\left [ \frac{\pi}{4}-\frac{1}{2} \right ]- \frac{n+1}2 \left [ I_{n+2}\;\;-\frac{1}{n+2} +I_n \right ]

\\I_{n+2}+\frac{n+1}2I_{n+2}+\frac{n+1}2 I_n= \frac{\pi}{4}-\frac{1}{2} +\frac{n+1}{2n+4}

\\\frac{n+3}2I_{n+2}+\frac{n+1}2 I_n= \frac{\pi}{4} +\frac{n+1-n-2}{2n+4}

\\\frac{n+3}2I_{n+2}+\frac{n+1}2 I_n= \frac{\pi}{4} -\frac{1}{2n+4}

\\{(n+3)}I_{n+2}+{(n+1)} I_n= \frac{\pi}{2} -\frac{1}{n+2}

Correct option (a)

Posted by

seema garhwal

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