JEE Mathematics Indefinite Integration MCQs Set 25

Answer: (c) 1

Explanation:
1. \( f'(x)=\frac{2\sin x-\sin2x}{x^3} \).
2. Using \( \sin x=x-\frac{x^3}{6}+\dots \) and \( \sin2x=2x-\frac{4x^3}{3}+\dots \), the numerator is \( 2x-\frac{x^3}{3}-2x+\frac{4x^3}{3}+\dots=x^3+\dots \).
3. Therefore \( \lim_{x\to0}f'(x)=1 \).

Teacher's Note:
1. The derivative of an antiderivative is the integrand, so only a limit is needed.

Answer: (c) \( 2e^{\sqrt x}\left[x-\sqrt x+1\right]+C \)

Explanation:
1. The integrand is \( e^{\sqrt x}\left(\sqrt x+1\right) \). Put \( t=\sqrt x \), so \( dx=2t\,dt \).
2. The integral becomes \( 2\int e^t\left(t^2+t\right)dt \). Since \( \frac{d}{dt}\left[e^t\left(t^2-t+1\right)\right]=e^t\left(t^2+t\right) \), it equals \( 2e^t\left(t^2-t+1\right) \).
3. So the value is \( 2e^{\sqrt x}\left(x-\sqrt x+1\right)+C \).

Teacher's Note:
1. After the substitution, integrate e^t times a polynomial by the e^t(f + f') idea.

Answer: (b) \( \ell n|x|-\frac27\ell n\left|1+x^7\right|+C \)

Explanation:
1. \( \frac{1-x^7}{x\left(1+x^7\right)}=\frac1x-\frac{2x^6}{1+x^7} \), since \( \frac1x-\frac{2x^6}{1+x^7}=\frac{1+x^7-2x^7}{x(1+x^7)} \).
2. \( \int\frac{2x^6}{1+x^7}dx=\frac27\ln\left|1+x^7\right| \).
3. So the value is \( \ln|x|-\frac27\ln\left|1+x^7\right|+C \).

Teacher's Note:
1. Split the fraction into 1/x minus a logarithmic derivative.

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

Explanation:
1. For \( x>1 \), \( \sqrt{\frac{x-1}{x+1}}\cdot\frac1{x^2}=\frac{x-1}{x^2\sqrt{x^2-1}} \).
2. \( \int\frac{dx}{x\sqrt{x^2-1}}=\sec^{-1}x \) and \( \frac{d}{dx}\frac{\sqrt{x^2-1}}{x}=\frac{1}{x^2\sqrt{x^2-1}} \).
3. Hence the integral is \( \sec^{-1}x-\frac{\sqrt{x^2-1}}{x}+C \).

Teacher's Note:
1. Multiply numerator and denominator by (x - 1) inside the root to remove the fraction under the root.

Answer: (b) \( \tan^{-1}\left(\tan^2x\right) \)

Explanation:
1. Divide the numerator and denominator by \( \cos^4x \): \( \frac{\sin2x}{\sin^4x+\cos^4x}=\frac{2\tan x\sec^2x}{1+\tan^4x} \).
2. Put \( t=\tan^2x \), so \( dt=2\tan x\sec^2x\,dx \).
3. The integral is \( \int\frac{dt}{1+t^2}=\tan^{-1}\left(\tan^2x\right)+C \). Option c differs by a sign (it equals \( \frac\pi2-\tan^{-1}(\tan^2x) \)).

Teacher's Note:
1. Dividing by cos^4 x turns sin 2x into 2 tan x sec^2 x, the derivative of tan^2 x.

Answer: (b) I and II only

Explanation:
1. \( \frac{d}{dx}\,2\sin^{-1}\sqrt x=\frac{2}{\sqrt{1-x}}\cdot\frac1{2\sqrt x}=\frac1{\sqrt{x-x^2}} \), so I is correct.
2. \( \frac{d}{dx}\sin^{-1}(2x-1)=\frac{2}{\sqrt{1-(2x-1)^2}}=\frac{2}{2\sqrt{x-x^2}}=\frac1{\sqrt{x-x^2}} \), so II is correct.
3. \( \frac{d}{dx}\left[-2\cos^{-1}(2x-1)\right]=\frac{2}{\sqrt{x-x^2}} \), which is twice the integrand, so III is wrong.

Teacher's Note:
1. Differentiate each candidate; two of them use the substitution x = sin^2 theta in disguise.

Answer: (a) \( \sqrt2\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt2}\right)+C \)

Explanation:
1. Put \( v=\sqrt{\tan x}-\sqrt{\cot x} \). Then \( v^2=\tan x+\cot x-2 \), so \( \tan x+\cot x=v^2+2 \).
2. From the given derivative, \( dv=\frac12\left(\sqrt{\tan x}+\sqrt{\cot x}\right)(v^2+2)\,dx \), so \( \left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx=\frac{2\,dv}{v^2+2} \).
3. \( I=\int\frac{2\,dv}{v^2+2}=\sqrt2\tan^{-1}\frac{v}{\sqrt2}+C \).

Teacher's Note:
1. The given derivative is exactly what is needed for the substitution v = sqrt(tan x) - sqrt(cot x).

Answer: (d) \( \sqrt2\sin^{-1}(\cos x-\sin x)+C \)

Explanation:
1. Put \( t=\sin x-\cos x \), so \( dt=(\sin x+\cos x)dx \) and \( \sin x\cos x=\frac{1-t^2}{2} \).
2. The integrand carries an extra minus sign, so \( I=-\sqrt2\int\frac{dt}{\sqrt{1-t^2}}=-\sqrt2\sin^{-1}(\sin x-\cos x)+C \).
3. Since \( -\sin^{-1}(u)=\sin^{-1}(-u) \), this equals \( \sqrt2\sin^{-1}(\cos x-\sin x)+C \).

Teacher's Note:
1. In the third quadrant sin x + cos x is negative, which flips the sign compared with the first quadrant.

Answer: (a) \( \frac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{(m+1)/m}+C \)

Explanation:
1. Since \( 2x^{2m}+3x^m+6=\frac{2x^{3m}+3x^{2m}+6x^m}{x^m} \), the second factor is \( \left(2x^{3m}+3x^{2m}+6x^m\right)^{1/m}\cdot\frac1x \).
2. Put \( t=2x^{3m}+3x^{2m}+6x^m \). Then \( dt=6m\left(x^{3m}+x^{2m}+x^m\right)\frac{dx}{x} \).
3. The integral is \( \frac{1}{6m}\int t^{1/m}dt=\frac{1}{6m}\cdot\frac{t^{(m+1)/m}}{(m+1)/m}=\frac{t^{(m+1)/m}}{6(m+1)}+C \).

Teacher's Note:
1. Multiply and divide by x so that the derivative of the combined polynomial appears.

Answer: (c) \( \frac{-1}{(\sec x+\tan x)^{11/2}}\left\{\frac1{11}+\frac17(\sec x+\tan x)^2\right\}+K \)

Explanation:
1. Put \( u=\sec x+\tan x \). Then \( du=u\sec x\,dx \), and \( \sec x=\frac12\left(u+\frac1u\right) \). So \( \sec^2x\,dx=\frac{\sec x\,du}{u}=\frac12\left(1+u^{-2}\right)du \).
2. \( \frac12\int\left(u^{-9/2}+u^{-13/2}\right)du=-\frac17u^{-7/2}-\frac1{11}u^{-11/2} \).
3. This equals \( -u^{-11/2}\left\{\frac1{11}+\frac{u^2}7\right\} \), option c.

Teacher's Note:
1. With u = sec x + tan x, express sec x in terms of u to convert the whole integral.

Answer: (a) \( \frac1n\log\left|\frac{x^n}{x^n+1}\right|+c \)

Explanation:
1. Multiply the numerator and denominator by \( x^{n-1} \): \( \frac{x^{n-1}}{x^n(x^n+1)} \). Put \( t=x^n \), \( dt=nx^{n-1}dx \).
2. \( \frac1n\int\frac{dt}{t(t+1)}=\frac1n\ln\left|\frac{t}{t+1}\right| \).
3. So the value is \( \frac1n\log\left|\frac{x^n}{x^n+1}\right|+c \).

Teacher's Note:
1. Multiplying by x^(n-1) makes the substitution t = x^n possible.

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

Explanation:
1. Write \( \sin x=\sin\left((x-\alpha)+\alpha\right)=\sin(x-\alpha)\cos\alpha+\cos(x-\alpha)\sin\alpha \).
2. Then \( \frac{\sin x}{\sin(x-\alpha)}=\cos\alpha+\sin\alpha\cot(x-\alpha) \).
3. Integrating gives \( x\cos\alpha+\sin\alpha\log\sin(x-\alpha)+c \), so \( (A,B)=(\cos\alpha,\sin\alpha) \).

Teacher's Note:
1. Express x as (x - alpha) + alpha to split the numerator.

Answer: (a) \( \frac1{\sqrt2}\log\left|\tan\left(\frac x2+\frac{3\pi}8\right)\right|+c \)

Explanation:
1. \( \cos x-\sin x=\sqrt2\cos\left(x+\frac\pi4\right) \), so the integral is \( \frac1{\sqrt2}\int\sec\left(x+\frac\pi4\right)dx \).
2. Using \( \int\sec\theta\,d\theta=\ln\left|\tan\left(\frac\theta2+\frac\pi4\right)\right| \) with \( \theta=x+\frac\pi4 \): the argument is \( \frac x2+\frac\pi8+\frac\pi4=\frac x2+\frac{3\pi}8 \).
3. The value is \( \frac1{\sqrt2}\log\left|\tan\left(\frac x2+\frac{3\pi}8\right)\right|+c \).

Teacher's Note:
1. Reduce cos x - sin x to a single cosine, then use the standard integral of sec.

Answer: (c) \( \frac{x}{(\log x)^2+1}+c \)

Explanation:
1. Let \( L=\log x \). Differentiate \( \frac{x}{1+L^2} \): \( \frac{(1+L^2)-x\cdot\frac{2L}{x}}{(1+L^2)^2}=\frac{1-2L+L^2}{(1+L^2)^2}=\frac{(L-1)^2}{(1+L^2)^2} \).
2. This is the integrand.
3. So the value is \( \frac{x}{(\log x)^2+1}+c \).

Teacher's Note:
1. Integrals of the form (f + f') e^x have a log analogue; here guess x/(1 + log^2 x) and differentiate to confirm.

Answer: (c) \( \frac12\log\tan\left(\frac x2+\frac\pi{12}\right)+c \)

Explanation:
1. \( \cos x+\sqrt3\sin x=2\cos\left(x-\frac\pi3\right) \), so the integral is \( \frac12\int\sec\left(x-\frac\pi3\right)dx \).
2. This equals \( \frac12\ln\left|\tan\left(\frac{x}{2}-\frac\pi6+\frac\pi4\right)\right| \).
3. Since \( -\frac\pi6+\frac\pi4=\frac\pi{12} \), the value is \( \frac12\log\tan\left(\frac x2+\frac\pi{12}\right)+c \).

Teacher's Note:
1. Use the R cos(x - phi) form: R = 2 and phi = pi/3.

Answer: (d) \( 2 \)

Explanation:
1. \( \frac{5\tan x}{\tan x-2}=\frac{5\sin x}{\sin x-2\cos x} \). Write \( 5\sin x=A(\sin x-2\cos x)+B(\cos x+2\sin x) \), where \( \cos x+2\sin x \) is the derivative of the denominator.
2. Comparing coefficients: \( A+2B=5 \) and \( -2A+B=0 \), so \( A=1,\ B=2 \).
3. The integral is \( x+2\ln|\sin x-2\cos x|+k \), so \( a=2 \).

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

Answer: (a) \( \frac13x^3\psi\left(x^3\right)-\int x^2\psi\left(x^3\right)dx+C \)

Explanation:
1. Put \( u=x^3 \), so \( x^5dx=\frac13u\,du \). The integral is \( \frac13\int u\,f(u)\,du \).
2. Integrating by parts: \( \frac13\left[u\psi(u)-\int\psi(u)du\right] \).
3. Back in \(x\): \( \int\psi(u)du=\int\psi\left(x^3\right)3x^2dx \), so the result is \( \frac13x^3\psi\left(x^3\right)-\int x^2\psi\left(x^3\right)dx+C \).

Teacher's Note:
1. Substitute u = x^3 first, then integrate by parts with f(u) as the derivative part.

Answer: (b) 2

Explanation:
1. Any two antiderivatives of the same function differ by a constant.
2. The difference at \( x=1 \) is 2, so the constant is 2.
3. Hence the difference at \( x=100 \) is also 2.

Teacher's Note:
1. Two antiderivatives of one function can differ only by a constant, so the gap never changes with x.

Answer: (b) 1

Explanation:
1. \( \log_{ex}e=\frac{1}{1+\ln x} \) and \( \log_{e^2x}e=\frac{1}{2+\ln x} \). Put \( L=\ln x \), \( dL=\frac{dx}x \).
2. \( \int\frac{dL}{(1+L)(2+L)}=\ln\frac{1+L}{2+L}+C \), so \( f(x)=\ln x \).
3. \( \lim_{x\to0}\frac{\ln(1+x)}{x}=1 \).

Teacher's Note:
1. Change the base of each logarithm to e and substitute L = ln x.

Answer: (b) 1

Explanation:
1. Write \( x=\frac12(2x-4)+2 \). Then \( \int\frac{x\,dx}{x^2-4x+8}=\frac12\ln\left(x^2-4x+8\right)+2\int\frac{dx}{(x-2)^2+4} \).
2. \( 2\int\frac{dx}{(x-2)^2+4}=2\cdot\frac12\tan^{-1}\frac{x-2}{2}=\tan^{-1}\frac{x-2}2 \), matching the given form.
3. So \( k=\frac12 \) and \( 2k=1 \).

Teacher's Note:
1. Split the numerator into a multiple of the derivative of the quadratic plus a constant.

Answer: (c) 10

Explanation:
1. \( \frac{d}{dx}\frac{x^5}{x^5+x+1}=\frac{5x^4\left(x^5+x+1\right)-x^5\left(5x^4+1\right)}{\left(x^5+x+1\right)^2}=\frac{4x^5+5x^4}{\left(x^5+x+1\right)^2} \).
2. So \( f(x)=\frac{x^5}{x^5+x+1}+C \), and \( f(0)=0 \) gives \( C=0 \).
3. \( f(1)=\frac13 \), hence \( 30f(1)=10 \).

Teacher's Note:
1. Divide numerator and denominator by x^10 to see a perfect derivative, or test x^5/(x^5 + x + 1).

Answer: (d) 1024

Explanation:
1. Repeatedly use \( 2\sin\theta\cos\theta=\sin2\theta \): \( \sin x\cos x\cos2x\cos4x\cos8x\cos16x=\frac{\sin32x}{32} \).
2. \( \int\frac{\sin32x}{32}dx=-\frac{\cos32x}{32\cdot32} \).
3. So \( k=1024 \).

Teacher's Note:
1. Each cosine doubles the angle in the sine, five times in all.

Answer: (c) 0

Explanation:
1. \( \int\frac{x\,dx}{\sqrt{1-9x^2}}=-\frac19\sqrt{1-9x^2} \), so \( A=-\frac19 \).
2. Put \( u=\cos^{-1}3x \), \( du=-\frac{3\,dx}{\sqrt{1-9x^2}} \). Then \( \int\frac{(\cos^{-1}3x)^2}{\sqrt{1-9x^2}}dx=-\frac13\int u^2du=-\frac19u^3 \), so \( B=-\frac19 \).
3. \( A-B=0 \).

Teacher's Note:
1. Treat the two terms of the numerator separately with different substitutions.

Answer: (c) 3

Explanation:
1. Let \( f(x)=1+bx+cx^2 \). The integral is rational only if the partial-fraction expansion has no \( \frac1x \) and \( \frac1{x+1} \) terms.
2. The coefficient of \( \frac1x \) is \( \frac{d}{dx}\left[\frac{f(x)}{(x+1)^3}\right]_{x=0}=f'(0)-3f(0)=b-3 \), so \( b=3 \). (The \( \frac{1}{x+1} \) coefficient is then also zero.)
3. Hence \( f'(0)=b=3 \).

Teacher's Note:
1. A rational antiderivative means no logarithm terms, i.e. zero residue at each simple pole.

Answer: (b) 2

Explanation:
1. \( \left(x-x^3\right)^{1/3}=x\left(\frac1{x^2}-1\right)^{1/3} \), so the integrand is \( x^{-3}\left(\frac1{x^2}-1\right)^{1/3} \).
2. Put \( u=\frac1{x^2}-1 \), \( du=-2x^{-3}dx \). The integral is \( -\frac12\int u^{1/3}du=-\frac38u^{4/3} \).
3. So \( a=-\frac38,\ b=\frac43 \), \( |ab|=\frac12 \) and \( \frac1{|ab|}=2 \).

Teacher's Note:
1. Factor x out of the cube root to expose the combination 1/x^2 - 1.