JEE Mathematics Indefinite Integrals MCQs Set 11

Answer: (c) \( x+2\sin x+\sin 2x+C \)

Explanation:
1. Use \( \sin5\theta/\sin\theta=1+2\cos2\theta+2\cos4\theta \) with \( \theta=x/2 \): the integrand is \( 1+2\cos x+2\cos 2x \).
2. Integrating gives \( x+2\sin x+\sin 2x+C \). Answer: (c)

Teacher's Note:
1. Write the ratio as a sum of cosines before integrating.

Answer: (d) \( \dfrac1{16} \)

Explanation:
1. Divide by \( (x^2+1)^2 \): \( \dfrac{dy}{dx}+\dfrac{2x}{x^2+1}y=\dfrac{1}{(x^2+1)^2} \). The integrating factor is \( x^2+1 \).
2. \( y(x^2+1)=\int\frac{dx}{x^2+1}=\tan^{-1}x+C \), and \( y(0)=0 \) gives \( C=0 \). So \( y(1)=\dfrac{\pi/4}{2}=\dfrac{\pi}{8} \).
3. \( \sqrt a\cdot\dfrac\pi8=\dfrac\pi{32} \) gives \( \sqrt a=\dfrac14 \), so \( a=\dfrac1{16} \). Answer: (d)

Teacher's Note:
1. Use the integrating factor method.

Answer: (b) \( -\dfrac1{2x^3} \)

Explanation:
1. Write \( (1+x^6)^{2/3}=x^4(x^{-6}+1)^{2/3} \), so the integral is \( \int\dfrac{dx}{x^7(1+x^{-6})^{2/3}} \). Put \( 1+x^{-6}=t^3 \), so \( -6x^{-7}dx=3t^2dt \).
2. The integral becomes \( -\dfrac12\int dt=-\dfrac t2+C=-\dfrac1{2}(1+x^{-6})^{1/3}+C=-\dfrac{1}{2x^2}(1+x^6)^{1/3}+C \).
3. Comparing with \( x f(x)(1+x^6)^{1/3} \) gives \( f(x)=-\dfrac{1}{2x^3} \). Answer: (b)

Teacher's Note:
1. Substitute for \( 1+x^{-6} \).

Answer: (b) \( -3\tan^{-1/3}x+C \)

Explanation:
1. The integrand is \( \dfrac{1}{\cos^{2/3}x\sin^{4/3}x}=\dfrac{\sec^2x}{\tan^{4/3}x} \) (divide numerator and denominator by \( \cos^{4/3}x \)).
2. With \( t=\tan x \): \( \int t^{-4/3}dt=-3t^{-1/3}+C=-3\tan^{-1/3}x+C \). Answer: (b)

Teacher's Note:
1. Rewrite everything in terms of \( \tan x \).

Answer: (b) \( \sec x+\tan x+\dfrac12 \)

Explanation:
1. Differentiate both sides: \( e^{\sec x}[\sec x\tan x f+\sec x\tan x+\sec^2x]=e^{\sec x}[\sec x\tan x f+f'] \).
2. Hence \( f'(x)=\sec x\tan x+\sec^2x \), so \( f(x)=\sec x+\tan x+C \), and the option with \( C=\tfrac12 \) is (b). Answer: (b)

Teacher's Note:
1. Differentiate both sides.

Answer: (c) \( A=\dfrac1{54} \) and \( f(x)=3(x-1) \)

Explanation:
1. Put \( x-1=3\tan\theta \): \( I=\dfrac1{27}\int\cos^2\theta\,d\theta=\dfrac1{54}(\theta+\sin\theta\cos\theta)+C \).
2. With \( \sin\theta\cos\theta=\dfrac{3(x-1)}{x^2-2x+10} \) this is \( \dfrac1{54}\left[\tan^{-1}\dfrac{x-1}3+\dfrac{3(x-1)}{x^2-2x+10}\right] \). Answer: (c)

Teacher's Note:
1. Trigonometric substitution on \( (x-1)^2+3^2 \).

Answer: (d) \( -\dfrac52 \)

Explanation:
1. Put \( t=x^2 \): \( \frac12\int t^2e^{-t}dt=-\frac12e^{-t}(t^2+2t+2)+C \).
2. So \( g(x)=-\dfrac12(x^4+2x^2+2) \) and \( g(-1)=-\dfrac12(1+2+2)=-\dfrac52 \). Answer: (d)

Teacher's Note:
1. Substitute \( t=x^2 \) and integrate by parts.

Answer: (c) \( \log_e\left|\dfrac{x^3+1}{x}\right|+C \)

Explanation:
1. Divide numerator and denominator by \( x^2 \): \( \dfrac{2x-1/x^2}{x^2+1/x} \).
2. Put \( t=x^2+\frac1x \), so \( dt=(2x-\frac1{x^2})dx \), and the integral is \( \log_e|t|+C=\log_e\left|\dfrac{x^3+1}{x}\right|+C \). Answer: (c)

Teacher's Note:
1. Divide by \( x^2 \) and look for a derivative in the numerator.

Answer: (b) \( x-\alpha \) and \( \log_e|\sin(x-\alpha)| \)

Explanation:
1. The integrand is \( \dfrac{\sin(x+\alpha)}{\sin(x-\alpha)} \). Put \( t=x-\alpha \): \( \int\dfrac{\sin(t+2\alpha)}{\sin t}dt=\int(\cos2\alpha+\sin2\alpha\cot t)\,dt \).
2. This is \( t\cos2\alpha+\sin2\alpha\log_e|\sin t|+C \), so \( A=x-\alpha \) and \( B=\log_e|\sin(x-\alpha)| \). Answer: (b)

Teacher's Note:
1. Combine the tangents into a ratio of sines.

Answer: (b) \( \log_e\left|\sec\left(\dfrac{x^2-1}2\right)\right|+C \)

Explanation:
1. Put \( \theta=\dfrac{x^2-1}2 \), so \( x\,dx=d\theta \). The ratio is \( \dfrac{2\sin2\theta-\sin4\theta}{2\sin2\theta+\sin4\theta}=\dfrac{1-\cos2\theta}{1+\cos2\theta}=\tan^2\theta \).
2. So the integral is \( \int|\tan\theta|\,d\theta=\log_e|\sec\theta|+C=\log_e\left|\sec\dfrac{x^2-1}2\right|+C \). Answer: (b)

Teacher's Note:
1. Substitute \( \theta=(x^2-1)/2 \).

Answer: (c) \( \dfrac14 \)

Explanation:
1. Divide numerator and denominator by \( x^{14} \): the integrand is \( \dfrac{5x^{-6}+7x^{-8}}{(x^{-5}+x^{-7}+2)^2} \), and the numerator is \( -\frac{d}{dx}(x^{-5}+x^{-7}+2) \).
2. So \( f(x)=\dfrac1{x^{-5}+x^{-7}+2}+C=\dfrac{x^7}{x^2+1+2x^7}+C \); \( f(0)=0 \) gives \( C=0 \) and \( f(1)=\dfrac14 \). Answer: (c)

Teacher's Note:
1. Divide through by \( x^{14} \).

Answer: (c) \( \dfrac n{n^2-1}\left(1-\dfrac1{\sin^{n-1}\theta}\right)^{\frac{n+1}n}+C \)

Explanation:
1. Put \( t=\sin\theta \): the integrand becomes \( \dfrac{(1-t^{1-n})^{1/n}}{t^n}\,dt \).
2. Let \( u=1-t^{1-n} \), so \( du=(n-1)t^{-n}dt \). Then \( \dfrac1{n-1}\int u^{1/n}du=\dfrac{n}{n^2-1}u^{\frac{n+1}n}+C \). Answer: (c)

Teacher's Note:
1. Substitute \( t=\sin\theta \), then \( u=1-t^{1-n} \).

Answer: (a) \( -4x^3-1 \)

Explanation:
1. Put \( t=x^3 \): \( \frac13\int te^{-4t}dt=\frac13\left[-\frac{te^{-4t}}4-\frac{e^{-4t}}{16}\right]=-\frac1{48}e^{-4t}(4t+1)+C \).
2. Comparing with \( \frac1{48}e^{-4x^3}f(x) \) gives \( f(x)=-(4x^3+1)=-4x^3-1 \). Answer: (a)

Teacher's Note:
1. Substitute \( t=x^3 \) and integrate by parts.

Answer: (c) \( \dfrac{-1}{27x^9} \)

Explanation:
1. Write \( \sqrt{1-x^2}=x\sqrt{x^{-2}-1} \) so the integrand is \( x^{-3}\sqrt{x^{-2}-1} \). Put \( x^{-2}-1=t^2 \), so \( x^{-3}dx=-t\,dt \).
2. The integral is \( -\int t^2dt=-\dfrac{t^3}3+C=-\dfrac1{3x^3}\left(\sqrt{1-x^2}\right)^3+C \), so \( A=-\dfrac1{3x^3} \), \( m=3 \) and \( A^m=-\dfrac1{27x^9} \). Answer: (c)

Teacher's Note:
1. Substitute \( t=\sqrt{x^{-2}-1} \).

Answer: (b) \( \dfrac13(x+4) \)

Explanation:
1. Put \( 2x-1=t^2 \): \( \int\dfrac{(t^2+3)/2}{t}\,t\,dt=\dfrac12\left(\dfrac{t^3}3+3t\right)+C=\dfrac t6(t^2+9)+C \).
2. This equals \( \dfrac{\sqrt{2x-1}}6(2x+8)=\dfrac{x+4}3\sqrt{2x-1} \), so \( f(x)=\dfrac{x+4}3 \). Answer: (b)

Teacher's Note:
1. Substitute \( t=\sqrt{2x-1} \).

Answer: (a) \( \dfrac x2\left[\cos(\log_ex)+\sin(\log_ex)\right]+C \)

Explanation:
1. Integrating by parts twice: \( I=x\cos(\ln x)+\int\sin(\ln x)\,dx \) and \( \int\sin(\ln x)dx=x\sin(\ln x)-I \).
2. So \( I=x\cos(\ln x)+x\sin(\ln x)-I \), giving \( I=\dfrac x2[\cos(\ln x)+\sin(\ln x)]+C \). Answer: (a)

Teacher's Note:
1. Integrate by parts twice and solve for \( I \).

Answer: (b) \( \dfrac{x^{12}}{6(2x^4+3x^2+1)^3}+C \)

Explanation:
1. Divide numerator and denominator by \( x^{16} \): the integrand is \( \dfrac{3x^{-3}+2x^{-5}}{(2+3x^{-2}+x^{-4})^4} \). Put \( t=2+3x^{-2}+x^{-4} \), so \( (3x^{-3}+2x^{-5})dx=-\dfrac{dt}2 \).
2. \( I=-\dfrac12\int t^{-4}dt=\dfrac1{6t^3}+C=\dfrac{x^{12}}{6(2x^4+3x^2+1)^3}+C \). Answer: (b)

Teacher's Note:
1. Divide through by \( x^{16} \).

Answer: (b) \( \dfrac{-1}{3(1+\tan^3x)}+C \)

Explanation:
1. The bracket is \( \sin^3x(\sin^2x+\cos^2x)+\cos^3x(\sin^2x+\cos^2x)=\sin^3x+\cos^3x \).
2. So the integrand is \( \dfrac{\sin^2x\cos^2x}{\cos^6x(1+\tan^3x)^2}=\dfrac{\tan^2x\sec^2x}{(1+\tan^3x)^2} \). With \( t=\tan^3x \): \( \dfrac13\int\dfrac{dt}{(1+t)^2}=-\dfrac1{3(1+\tan^3x)}+C \). Answer: (b)

Teacher's Note:
1. Factor the long bracket first.

Answer: (b) \( \dfrac{x^{10}}{2(x^5+x^3+1)^2}+C \)

Explanation:
1. Divide by \( x^{15} \): \( \dfrac{2x^{-3}+5x^{-6}}{(1+x^{-2}+x^{-5})^3} \). Put \( t=1+x^{-2}+x^{-5} \), so \( (2x^{-3}+5x^{-6})dx=-dt \).
2. \( I=-\int t^{-3}dt=\dfrac1{2t^2}+C=\dfrac{x^{10}}{2(x^5+x^3+1)^2}+C \). Answer: (b)

Teacher's Note:
1. Divide numerator and denominator by \( x^{15} \).

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

Explanation:
1. Write \( x^2(x^4+1)^{3/4}=x^5\left(1+x^{-4}\right)^{3/4} \). Put \( 1+x^{-4}=t^4 \), so \( -4x^{-5}dx=4t^3dt \).
2. \( I=-\int dt=-t+c=-\left(1+\dfrac1{x^4}\right)^{1/4}+c \). Answer: (d)

Teacher's Note:
1. Factor \( x^4 \) out of the bracket.

Answer: (b) \( x\,e^{x+\frac1x}+C \)

Explanation:
1. Split the integrand: \( e^{x+1/x}+x\left(1-\dfrac1{x^2}\right)e^{x+1/x} \). Note \( \dfrac{d}{dx}e^{x+1/x}=\left(1-\dfrac1{x^2}\right)e^{x+1/x} \).
2. Integrating the second part by parts gives \( xe^{x+1/x}-\int e^{x+1/x}dx \), so the total is \( xe^{x+1/x}+C \). Answer: (b)

Teacher's Note:
1. Split into \( e^{f} \) and \( x f'e^{f} \) and use parts.

Answer: (c) \( \dfrac13x^3\psi(x^3)-\displaystyle\int x^2\psi(x^3)\,dx+C \)

Explanation:
1. Put \( t=x^3 \), \( x^2dx=\dfrac{dt}3 \): \( I=\dfrac13\int t\,f(t)\,dt=\dfrac13\left[t\psi(t)-\int\psi(t)\,dt\right] \).
2. Returning to \( x \), \( \int\psi(t)dt=3\int x^2\psi(x^3)dx \), so \( I=\dfrac13x^3\psi(x^3)-\int x^2\psi(x^3)\,dx+C \). Answer: (c)

Teacher's Note:
1. Substitute \( t=x^3 \) and integrate by parts.

Answer: (d) \( 2 \)

Explanation:
1. The integrand is \( \dfrac{5\sin x}{\sin x-2\cos x} \). Write \( 5\sin x=1\cdot(\sin x-2\cos x)+2(\cos x+2\sin x) \).
2. So the integral is \( \int1\,dx+2\int\dfrac{\cos x+2\sin x}{\sin x-2\cos x}dx=x+2\log|\sin x-2\cos x|+k \), hence \( a=2 \). Answer: (d)

Teacher's Note:
1. Write the numerator as a combination of the denominator and its derivative.

Answer: (d) \( x+\log\left|\sin\left(x-\frac\pi4\right)\right|+C \)

Explanation:
1. Put \( t=x-\dfrac\pi4 \): \( \sqrt2\sin\left(t+\dfrac\pi4\right)=\sin t+\cos t \), so the integral is \( \int\dfrac{\sin t+\cos t}{\sin t}dt=\int(1+\cot t)\,dt \).
2. This is \( t+\log|\sin t|+C' \), i.e. \( x+\log\left|\sin\left(x-\frac\pi4\right)\right|+C \). Answer: (d)

Teacher's Note:
1. Substitute \( t=x-\pi/4 \).

Answer: (a) \( \dfrac12\log\tan\left(\dfrac x2+\dfrac\pi{12}\right)+C \)

Explanation:
1. \( \cos x+\sqrt3\sin x=2\cos\left(x-\dfrac\pi3\right) \), so the integral is \( \dfrac12\int\sec\left(x-\dfrac\pi3\right)dx \).
2. \( \int\sec u\,du=\log\tan\left(\dfrac u2+\dfrac\pi4\right) \) gives \( \dfrac12\log\tan\left(\dfrac x2-\dfrac\pi6+\dfrac\pi4\right)=\dfrac12\log\tan\left(\dfrac x2+\dfrac\pi{12}\right)+C \). Answer: (a)

Teacher's Note:
1. Write the denominator as a single cosine.

Answer: (a) \( \dfrac x{(\log x)^2+1}+C \)

Explanation:
1. The integrand is \( \dfrac{(\log x)^2+1-2\log x}{[(\log x)^2+1]^2}=\dfrac1{(\log x)^2+1}-\dfrac{2\log x}{[(\log x)^2+1]^2} \).
2. This is exactly \( \dfrac{d}{dx}\left[\dfrac x{(\log x)^2+1}\right] \), since the derivative is \( \dfrac1{(\log x)^2+1}-\dfrac{2\log x}{[(\log x)^2+1]^2} \). Answer: (a)

Teacher's Note:
1. Expand the numerator, then guess the antiderivative of the form \( x/g(x) \).

Answer: (b) \( (\cos\alpha,\sin\alpha) \)

Explanation:
1. Put \( t=x-\alpha \): \( \dfrac{\sin(t+\alpha)}{\sin t}=\cos\alpha+\sin\alpha\cot t \).
2. The integral is \( t\cos\alpha+\sin\alpha\log\sin t+C' \), so \( A=\cos\alpha \) and \( B=\sin\alpha \). Answer: (b)

Teacher's Note:
1. Substitute \( t=x-\alpha \) and expand \( \sin(t+\alpha) \).

Answer: (d) \( \dfrac1{\sqrt2}\log\left|\tan\left(\dfrac x2+\dfrac{3\pi}8\right)\right|+C \)

Explanation:
1. \( \cos x-\sin x=\sqrt2\cos\left(x+\dfrac\pi4\right) \), so the integral is \( \dfrac1{\sqrt2}\int\sec\left(x+\dfrac\pi4\right)dx \).
2. This equals \( \dfrac1{\sqrt2}\log\left|\tan\left(\dfrac x2+\dfrac\pi8+\dfrac\pi4\right)\right|=\dfrac1{\sqrt2}\log\left|\tan\left(\dfrac x2+\dfrac{3\pi}8\right)\right|+C \). Answer: (d)

Teacher's Note:
1. Write the denominator as a single cosine.

Answer: (a) \( \dfrac1n\log\left(\dfrac{x^n}{x^n+1}\right)+C \)

Explanation:
1. Multiply numerator and denominator by \( x^{n-1} \) and put \( t=x^n+1 \), \( nx^{n-1}dx=dt \).
2. \( I=\dfrac1n\int\dfrac{dt}{t(t-1)}=\dfrac1n\log\dfrac{t-1}t+C=\dfrac1n\log\dfrac{x^n}{x^n+1}+C \). Answer: (a)

Teacher's Note:
1. Make the numerator the derivative of \( x^n \).