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Choose the correct answers in Exercises 41 to 44.

Q44.    The value of $\int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx$ is

(A)    1

(B)    0

(C)    -1

(D)    $\frac{\pi}{4}$

as       Now the integral can be written as (B) is correct.

Choose the correct answers in Exercises 41 to 44.

Q43.    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)    $\frac{a+b}{2}\int^b_af(b+x)dx$

(C)    $\frac{b-a}{2}\int^b_af(x)dx$

(D)    $\frac{a+b}{2}\int^b_af(x)dx$

As we know  Using the above property we can write the integral as Answer (D) is correct

Choose the correct answers in Exercises 41 to 44.

Q42.    $\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx$ is equal to

(A)    $\frac{-1}{\sin x + \cos x} + C$

(B)    $\log |{\sin x + \cos x} |+ C$

(C)    $\log |{\sin x- \cos x} |+ C$

(D)    $\frac{1}{(\sin x + \cos x)^2} + C$

cos2x=cos2x-sin2x let sinx+cosx=t,(cosx-sinx)dx=dt hence the given integral can be written as B is correct

Choose the correct answers in Exercises 41 to 44.

Q41.    $\int\frac{dx}{e^x + e^{-x}}$ is equal to

(A)    $\tan^{-1}(e^x) + c$

(B)    $\tan^{-1}(e^{-x}) + c$

(C)    $\log (e^x - e^{-x}) + C$

(D)    $\log (e^x + e^{-x}) + C$

the above integral can be re arranged as let ex=t, exdx=dt   (A) is correct

Q40.    Evaluate $\int_0^1e^{2-3x}dx$as a limit of a sum.

As we know where b-a=hn In the given problem b=1, a=0 and

Prove the following (Exercises 34 to 39)\

Q39.    $\int_0^1\sin^{-1}xdx = \frac{\pi}{2}-1$

Integrating by parts we get For I2 take 1-x2 = t2, -xdx=tdt Hence Proved

Prove the following (Exercises 34 to 39)

Q38.    $\int_0^\frac{\pi}{4}2\tan^3 x dx = 1 - \log 2$

The integral is written as Hence Proved

Prove the following (Exercises 34 to 39)

Q37.    $\int_0^\frac{\pi}{2}\sin^3 x dx =\frac{2}{3}$

For I2 let cosx=t, -sinxdx=dt The limits change to 0 and 1 I1-I2=2/3 Hence proved.

Prove the following (Exercises 34 to 39)

Q36.    $\int_{-1}^1x^{17}\cos^4 x dx=0$

The Integrand g(x) therefore is an odd function and therefore

Prove the following (Exercises 34 to 39)

Q34.    $\int_1^3\frac{dx}{x^2(x+1)} = \frac{2}{3}+ \log \frac{2}{3}$

L.H.S =  We can write the numerator as [(x+1) -x]       = RHS Hence proved.

Evaluate the definite integrals in Exercises 25 to 33.

Q33.    $\int_1^4[|x-1| + |x-2| + |x-3|]dx$

Given integral  So, we split it in according to intervals they are positive or negative. Now,   as  is positive in the given x -range                       Therefore,    as  is in the given x -range  and  in the range  Therefore,    as  is in the given x -range  and  in the range  Therefore,  So, We have the sum

Evaluate the definite integrals in Exercises 25 to 33.

Q32.    $\int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx$

Let I =                -(i) Replacing x with ( -x),                    - (ii) Adding (i) and (ii)

Evaluate the definite integrals in Exercises 25 to 33.

Q31.    $\int_0^\frac{\pi}{2}\sin 2x\tan^{-1}(\sin x)dx$

Let I = 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 Now the important step here is to change the limit of the integration as we are changing the variable.so, So our function becomes, Now, let's integrate this by using integration by parts method,

Evaluate the definite integrals in Exercises 25 to 33.

Q30.    $\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx$

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   Now, Hence our function in terms of t becomes,

Evaluate the definite integrals in Exercises 25 to 33.

Q29.    $\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}$

First, let's get rid of the square roots from the denominator,

Evaluate the definite integrals in Exercises 25 to 33.

Q28.    $\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}$

Here first let convert sin2x as the angle of x ( sinx, and cosx)  Now let's remove the square root form function by making a perfect square inside the square root Now let ,  since we are changing the variable, limit of integration will change our function in terms of t :

Evaluate the definite integrals in Exercises 25 to 33.

Q27.    $\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}$

Lets first simplify the function.   As we have a good relation in between squares of the tan and square of sec lets try to take our equation there, AS we can write square of sec in term of tan,   Now let's calculate the integral of the second function, (we already have calculated the first function) let   here we are changing the variable so we have to calculate the limits of the new...

Evaluate the definite integrals in Exercises 25 to 33.

Q26.    $\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}$

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  Now lets change the variable the limits will also change since the variable is changing  So, the integration becomes:

Evaluate the definite integrals in Exercises 25 to 33.

Q25.    $\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx$

Since, we have   multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,   So, Here let's use the property so,

Integrate the functions in Exercises 1 to 24.

Q24.    $\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}$

Here let's first reduce the log function. Now, let  So our function in terms if new variable t is : now let's solve this By using integration by parts
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