JEE Mathematics Indefinite Integration MCQs Set 24

Answer: (b) \( \frac{16}{5} \)

Explanation:
1. Since \( \sqrt{2\sin2x}=2\sqrt{\sin x\cos x} \) and \( \sin x\cos x=\tan x\cos^2x \), the integrand becomes \( \frac{\sec^4x}{2\sqrt{\tan x}} \).
2. Put \( t=\tan x \), so \( \sec^4x\,dx=(1+t^2)\,dt \). The integral is \( \frac12\int\left(t^{-1/2}+t^{3/2}\right)dt=t^{1/2}+\frac15t^{5/2} \).
3. Hence \( A=\frac12,\ C=\frac15,\ B=\frac52 \) and \( A+B+C=\frac{16}{5} \).

Teacher's Note:
1. Rewrite the root in terms of tan x so that sec^2 x dx becomes dt.

Answer: (a) \( -2\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}+C \)

Explanation:
1. Put \( \sqrt x=\sin\theta \). Then \( dx=2\sin\theta\cos\theta\,d\theta \) and \( \sqrt{x-x^2}=\sin\theta\cos\theta \).
2. The integral becomes \( \int\frac{2\,d\theta}{1+\sin\theta}=2(\tan\theta-\sec\theta)=-2\sqrt{\frac{1-\sin\theta}{1+\sin\theta}} \).
3. Replacing \( \sin\theta=\sqrt x \) gives \( -2\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}+C \).

Teacher's Note:
1. A sine substitution removes the root; the final result is easiest to read back in terms of sqrt(x).

Answer: (d) \( -\frac43\left(\frac{x+1}{x-2}\right)^{1/4}+C \)

Explanation:
1. Write the integrand as \( \frac{1}{(x-2)^2}\left(\frac{x+1}{x-2}\right)^{-3/4} \).
2. Put \( u=\frac{x+1}{x-2} \), so \( du=-\frac{3}{(x-2)^2}dx \).
3. The integral is \( -\frac13\int u^{-3/4}du=-\frac13\cdot4u^{1/4}=-\frac43\left(\frac{x+1}{x-2}\right)^{1/4}+C \).

Teacher's Note:
1. When the exponents add up to -2 for factors (x-a)^p (x-b)^q, substitute u = (x-a)/(x-b).

Answer: (d) \( \log\left(2+\sqrt5\right) \)

Explanation:
1. Put \( u=\log\left(t+\sqrt{1+t^2}\right) \). Then \( du=\frac{dt}{\sqrt{1+t^2}} \).
2. The integral becomes \( \int u\,du=\frac{u^2}{2}+C \), so \( g(t)=\log\left(t+\sqrt{1+t^2}\right) \).
3. Therefore \( g(2)=\log\left(2+\sqrt5\right) \).

Teacher's Note:
1. The derivative of log(t + sqrt(1+t^2)) is exactly 1/sqrt(1+t^2), which is the other factor in the integrand.

Answer: (c) \( -\frac12\sin2x+c \)

Explanation:
1. \( \sin^8x-\cos^8x=(\sin^4x-\cos^4x)(\sin^4x+\cos^4x)=(\sin^2x-\cos^2x)(\sin^4x+\cos^4x) \).
2. Also \( 1-2\sin^2x\cos^2x=\sin^4x+\cos^4x \). So the integrand reduces to \( \sin^2x-\cos^2x=-\cos2x \).
3. \( \int-\cos2x\,dx=-\frac12\sin2x+c \).

Teacher's Note:
1. Factorise the numerator using a^2 - b^2 repeatedly and compare with the denominator.

Answer: (c) \( -\frac{1}{3\left(1+\tan^3x\right)}+c \)

Explanation:
1. Divide the numerator and denominator by \( \cos^6x \): the integrand becomes \( \frac{\tan^2x\sec^2x}{\left(1+\tan^3x\right)^2} \).
2. Put \( t=1+\tan^3x \), so \( dt=3\tan^2x\sec^2x\,dx \).
3. \( \frac13\int\frac{dt}{t^2}=-\frac{1}{3t}=-\frac{1}{3\left(1+\tan^3x\right)}+c \).

Teacher's Note:
1. Dividing by cos^6 x converts everything into tan x and the single factor sec^2 x for dx.

Answer: (c) \( \ln|x|-p(x)+c \)

Explanation:
1. Note that \( \frac{1}{x+x^7}+\frac{x^6}{x+x^7}=\frac{1+x^6}{x(1+x^6)}=\frac1x \).
2. So \( \frac{x^6}{x+x^7}=\frac1x-\frac{1}{x+x^7} \).
3. Integrating, \( \int\frac{x^6}{x+x^7}dx=\ln|x|-p(x)+c \).

Teacher's Note:
1. Add the two integrands; the sum is simply 1/x.

Answer: (a) \( x \)

Explanation:
1. Differentiate \( x\,e^{\cot^{-1}x} \): the derivative is \( e^{\cot^{-1}x}\left(1-\frac{x}{1+x^2}\right) \).
2. \( 1-\frac{x}{1+x^2}=\frac{x^2-x+1}{x^2+1} \), which is the given integrand's coefficient.
3. Hence \( A(x)=x \).

Teacher's Note:
1. Guess a product of the form A(x) e^(cot^-1 x) and verify by differentiating.

Answer: (c) \( -\log\left|1+\sqrt{2-x^2}\right|+c \)

Explanation:
1. Put \( u=\sqrt{2-x^2} \). Then \( x\,dx=-u\,du \) and the denominator is \( u^2+u \).
2. The integral becomes \( -\int\frac{u\,du}{u(u+1)}=-\int\frac{du}{u+1}=-\log|1+u|+c \).
3. Therefore the result is \( -\log\left|1+\sqrt{2-x^2}\right|+c \).

Teacher's Note:
1. Substitute the square root itself; x dx supplies exactly the factor needed.

Answer: (b) \( 2\log\left|1+\sqrt x\right|+C \)

Explanation:
1. Put \( t=\sqrt x \), so \( dx=2t\,dt \).
2. The integral becomes \( \int\frac{2t\,dt}{t(1+t)}=2\log|1+t|+C \).
3. Hence the value is \( 2\log\left|1+\sqrt x\right|+C \).

Teacher's Note:
1. Substituting t = sqrt(x) cancels the root in the denominator.

Answer: (a) \( \log|x+\log\sin x|+C \)

Explanation:
1. The derivative of \( x+\log\sin x \) is \( 1+\cot x \).
2. So the integral has the form \( \int\frac{f'(x)}{f(x)}dx=\log|f(x)|+C \).
3. The value is \( \log|x+\log\sin x|+C \).

Teacher's Note:
1. Check whether the numerator is the derivative of the denominator.

Answer: (a) \( \frac{\left[\log(\sin x)\right]^2}{2}+C \)

Explanation:
1. Write the integrand as \( \log(\sin x)\cot x \).
2. Put \( u=\log\sin x \), so \( du=\cot x\,dx \).
3. \( \int u\,du=\frac{u^2}{2}+C=\frac{\left[\log(\sin x)\right]^2}{2}+C \).

Teacher's Note:
1. cot x dx is the differential of log(sin x).

Answer: (a) \( \frac13\left(x^2+x\right)^{3/2}-\frac14\left[\left(x+\frac12\right)\sqrt{x^2+x}-\frac14\log\left(x+\frac12+\sqrt{x^2+x}\right)\right]+c \)

Explanation:
1. Write \( x=\frac12(2x+1)-\frac12 \). Then \( \int x\sqrt{x^2+x}\,dx=\frac12\int(2x+1)\sqrt{x^2+x}\,dx-\frac12\int\sqrt{x^2+x}\,dx \).
2. The first part is \( \frac13\left(x^2+x\right)^{3/2} \). The second uses \( \int\sqrt{t^2-a^2}\,dt=\frac t2\sqrt{t^2-a^2}-\frac{a^2}{2}\log\left|t+\sqrt{t^2-a^2}\right| \) with \( t=x+\frac12,\ a=\frac12 \).
3. Combining gives option a, and differentiating it returns \( x\sqrt{x^2+x} \).

Teacher's Note:
1. Split x into a multiple of the derivative of x^2 + x plus a constant.

Answer: (d) \( \operatorname{cosec}\alpha\,\ln\left|\frac{\sin x}{\sin(x+\alpha)}\right|+C \)

Explanation:
1. Since \( \sin\alpha=\sin((x+\alpha)-x) \), we get \( \cot x-\cot(x+\alpha)=\frac{\sin\alpha}{\sin x\sin(x+\alpha)} \).
2. So \( \frac{1}{\sin x\sin(x+\alpha)}=\operatorname{cosec}\alpha\left[\cot x-\cot(x+\alpha)\right] \).
3. Integrating gives \( \operatorname{cosec}\alpha\left[\ln|\sin x|-\ln|\sin(x+\alpha)|\right]+C \).

Teacher's Note:
1. Write sin(alpha) as sin((x + alpha) - x) to split the product into a difference of cotangents.

Answer: (b) \( a=-\frac{5\pi}{4},\ b\in R \)

Explanation:
1. \( \int(\sin2x-\cos2x)dx=-\frac12\cos2x-\frac12\sin2x=-\frac{1}{\sqrt2}\sin\left(2x+\frac\pi4\right) \).
2. Since \( -\sin\theta=\sin(\theta+\pi) \), this equals \( \frac{1}{\sqrt2}\sin\left(2x+\frac{5\pi}{4}\right) \).
3. Comparing with \( \sin(2x-a) \) gives \( a=-\frac{5\pi}{4} \) with arbitrary constant \( b \).

Teacher's Note:
1. Combine -cos 2x/2 - sin 2x/2 into a single sine with amplitude 1/sqrt(2).

Answer: (a) \( 2\sin x-\ln|\sec x+\tan x|+C \)

Explanation:
1. \( \cos2x=2\cos^2x-1 \), so \( \frac{\cos2x}{\cos x}=2\cos x-\sec x \).
2. \( \int2\cos x\,dx=2\sin x \) and \( \int\sec x\,dx=\ln|\sec x+\tan x| \).
3. Hence the value is \( 2\sin x-\ln|\sec x+\tan x|+C \).

Teacher's Note:
1. Use the double-angle identity in the form 2cos^2 x - 1.

Answer: (a) \( 2\sqrt{\tan x}+C \)

Explanation:
1. Divide the numerator and denominator by \( \cos^2x \): \( \frac{1}{\sin x\cos x}=\frac{\sec^2x}{\tan x} \).
2. The integral becomes \( \int\frac{\sec^2x\,dx}{\sqrt{\tan x}} \). Put \( t=\tan x \): \( \int t^{-1/2}dt=2\sqrt t \).
3. So the value is \( 2\sqrt{\tan x}+C \).

Teacher's Note:
1. 1/(sin x cos x) = sec^2 x / tan x is a standard rewrite.

Answer: (d) \( -\left(1+\frac{1}{x^4}\right)^{1/4}+C \)

Explanation:
1. Take \( x^3 \) out of the bracket: \( \left(x^4+1\right)^{3/4}=x^3\left(1+x^{-4}\right)^{3/4} \), so the integrand is \( x^{-5}\left(1+x^{-4}\right)^{-3/4} \).
2. Put \( u=1+x^{-4} \), so \( du=-4x^{-5}dx \).
3. \( -\frac14\int u^{-3/4}du=-u^{1/4}=-\left(1+\frac1{x^4}\right)^{1/4}+C \).

Teacher's Note:
1. Factor out the highest power of x so that the bracket takes the form 1 + x^(-4).

Answer: (c) \( A=-\frac12,\ B=1 \)

Explanation:
1. Partial fractions: \( \frac{1}{x^3(x+1)}=\frac1{x^3}-\frac1{x^2}+\frac1x-\frac1{x+1} \).
2. Integrating term by term: \( -\frac1{2x^2}+\frac1x+\ln|x|-\ln|x+1| \).
3. Hence \( A=-\frac12 \) and \( B=1 \).

Teacher's Note:
1. Use partial fractions with the repeated factor x^3.

Answer: (c) \( -2\ell n\left(\cos\frac x2+\sqrt{\cos^2\frac x2-\frac12}\right)+C \)

Explanation:
1. \( \sec x-1=\frac{1-\cos x}{\cos x}=\frac{2\sin^2(x/2)}{2\cos^2(x/2)-1} \). Put \( t=\cos\frac x2 \), so \( \sin\frac x2\,dx=-2\,dt \).
2. The integral becomes \( -2\sqrt2\int\frac{dt}{\sqrt{2t^2-1}}=-2\int\frac{dt}{\sqrt{t^2-\frac12}} \).
3. \( =-2\ln\left(t+\sqrt{t^2-\frac12}\right)+C \).

Teacher's Note:
1. Half-angle identities turn both 1 - cos x and cos x into functions of cos(x/2).

Answer: (b) I, III and IV only

Explanation:
1. \( \frac{d}{dx}\left(x^3\sin\frac1x\right)=3x^2\sin\frac1x-x\cos\frac1x \), so \( f(x)=x^3\sin\frac1x+k \). The condition \( f(0)=0 \) gives \( k=0 \), consistent with \( f\left(\frac1\pi\right)=0 \).
2. \( f'(0)=\lim_{h\to0}h^2\sin\frac1h=0 \), so \( f \) is derivable, hence continuous, at 0. Statement I is true and II is false.
3. As \( x\to0 \), \( f'(x)\to0=f'(0) \), so \( f' \) is continuous at 0 and III is true.
4. \( \frac{f'(h)-f'(0)}{h}=3h\sin\frac1h-\cos\frac1h \) has no limit as \( h\to0 \), so \( f' \) is not derivable at 0 and IV is true. Hence I, III and IV are correct.

Teacher's Note:
1. First recover f by recognising the derivative of x^3 sin(1/x), then test each statement at 0.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. \( \log_xe=\frac{1}{\ln x} \). Put \( x=e^t \), so \( dx=e^tdt \).
2. The integral becomes \( \int\left(\frac1t-\frac1{t^2}\right)e^tdt \). With \( f(t)=\frac1t \) and \( f'(t)=-\frac1{t^2} \), the Reason gives \( e^t\cdot\frac1t+C=\frac{x}{\ln x}+C=x\log_xe+C \).
3. Both statements are true and the Reason is the method that proves the Assertion.

Teacher's Note:
1. The substitution x = e^t reduces the Assertion to the standard form e^t(f + f').

Answer: (a) \( -\ell n|\sec x|+\frac13\ell n|\sec3x|-\frac12\ell n|\sec2x|+C \)

Explanation:
1. From \( \tan3x=\frac{\tan x+\tan2x}{1-\tan x\tan2x} \) we get \( \tan x\tan2x\tan3x=\tan3x-\tan2x-\tan x \).
2. Using \( \int\tan kx\,dx=\frac1k\ell n|\sec kx| \), the integral is \( \frac13\ell n|\sec3x|-\frac12\ell n|\sec2x|-\ell n|\sec x|+C \).
3. This matches option a.

Teacher's Note:
1. Write 3x as x + 2x and use the tangent addition formula to split the product.

Answer: (c) \( e^{\sin x}(x-\sec x)+C \)

Explanation:
1. \( \frac{x\cos^3x-\sin x}{\cos^2x}=x\cos x-\sec x\tan x \).
2. Differentiate \( e^{\sin x}(x-\sec x) \): \( e^{\sin x}\cos x(x-\sec x)+e^{\sin x}(1-\sec x\tan x)=e^{\sin x}(x\cos x-\sec x\tan x) \).
3. This equals the integrand, so the value is \( e^{\sin x}(x-\sec x)+C \).

Teacher's Note:
1. Separate the integrand as e^(sin x) [f(x) cos x + f'(x)] with f(x) = x - sec x.

Answer: (b) \( \ell n\left|xe^{\sin x}\right|-\frac12\log\left|1-x^2e^{2\sin x}\right|+K \)

Explanation:
1. Put \( u=xe^{\sin x} \). Then \( du=e^{\sin x}(1+x\cos x)dx \), and the integrand is \( \frac{du}{u\left(1-u^2\right)} \).
2. \( \frac{1}{u(1-u^2)}=\frac1u+\frac{u}{1-u^2} \), so the integral is \( \ln|u|-\frac12\ln\left|1-u^2\right|+K \).
3. Substituting back gives option b.

Teacher's Note:
1. Notice that e^(sin x)(1 + x cos x) is the derivative of x e^(sin x).