JEE Mathematics Indefinite Integration MCQs Set 26

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

Explanation:
1. Use the power rule: \( \int u^{1/2}du=\frac23u^{3/2} \) with \( u=x+1 \).
2. So the result is \( \frac{2(x+1)^{3/2}}{3}+C \).

Teacher's Note:
1. Raise the power by 1 and divide by the new power.
2. The inner function x + 1 has derivative 1.

Answer: (b) \( \sqrt{2x+1}+C \)

Explanation:
1. Put \( u=2x+1 \), \( dx=du/2 \).
2. Then \( \frac12\int u^{-1/2}du=\sqrt u \), so the answer is \( \sqrt{2x+1}+C \).

Teacher's Note:
1. The derivative of 2x + 1 is 2, which cancels the factor 1/2.
2. Check by differentiating the answer.

Answer: (a) \( -\frac\pi4 \)

Explanation:
1. Write \( 1+\sin x=2\sin^2\left(\frac x2+\frac\pi4\right) \).
2. Then \( \int\frac{dx}{2\sin^2(\frac x2+\frac\pi4)}=-\cot\left(\frac x2+\frac\pi4\right) \).
3. Since \( -\cot\theta=\tan\left(\theta-\frac\pi2\right) \), the result is \( \tan\left(\frac x2-\frac\pi4\right) \), so \( a=-\frac\pi4 \).

Teacher's Note:
1. Use the half-angle identity for 1 + sin x.
2. Convert -cot into tan with a shift of pi/2.

Answer: (c) \( -\frac{5\pi}{4} \)

Explanation:
1. Integrate: \( -\frac12\cos2x-\frac12\sin2x \).
2. Write this as \( -\frac12(\sin2x+\cos2x)=-\frac1{\sqrt2}\sin\left(2x+\frac\pi4\right) \).
3. Now \( -\sin\theta=\sin(\theta+\pi) \), so it equals \( \frac1{\sqrt2}\sin\left(2x+\frac{5\pi}4\right) \) and \( -a=\frac{5\pi}4 \).

Teacher's Note:
1. Combine sin 2x + cos 2x into a single sine.
2. Note the sign in sin(2x - a).

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

Explanation:
1. Write \( \cos2x=2\cos^2x-1 \), so the integrand is \( 2\cos x-\sec x \).
2. Integrate: \( 2\sin x-\ln|\sec x+\tan x|+C \).

Teacher's Note:
1. Split the fraction after using the double-angle formula.
2. The integral of sec x is ln|sec x + tan x|.

Answer: (d) \( A=-\frac18,\ B\text{ arbitrary} \)

Explanation:
1. Use \( \cot x-\tan x=2\cot2x \) and \( 1+\cos4x=2\cos^22x \).
2. The integrand becomes \( \cos^22x\tan2x=\sin2x\cos2x=\tfrac12\sin4x \).
3. Integrating gives \( -\frac18\cos4x+B \).

Teacher's Note:
1. Simplify using double-angle identities first.
2. The constant of integration is the arbitrary B.

Answer: (b) \( \frac{2a^{\sqrt x}}{\ln a}+C \)

Explanation:
1. Put \( t=\sqrt x \), \( dt=\frac{dx}{2\sqrt x} \).
2. Then \( 2\int a^tdt=\frac{2a^t}{\ln a} \).

Teacher's Note:
1. The substitution t = sqrt x absorbs the 1/sqrt x.
2. The integral of a^t is a^t / ln a.

Answer: (d) \( \frac{5^{5^{5^x}}}{(\ln5)^3}+C \)

Explanation:
1. Let \( u=5^x \), \( v=5^u \), \( w=5^v \). Then \( du=u\ln5\,dx \), \( dv=v\ln5\,du \), \( dw=w\ln5\,dv \).
2. The integrand times dx equals \( w\,v\,u\,dx=\frac{w\,v\,du}{\ln5}=\frac{w\,dv}{(\ln5)^2}=\frac{dw}{(\ln5)^3} \).
3. Integrating gives \( \frac{5^{5^{5^x}}}{(\ln5)^3}+C \).

Teacher's Note:
1. Peel the nested exponentials one at a time.
2. Each step contributes one factor of 1/ln 5.

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

Explanation:
1. Divide numerator and denominator by \( \cos^2x \): \( \frac{\sqrt{\tan x}\,\sec^2x}{\tan x} \).
2. Put \( t=\tan x \): \( \int t^{-1/2}dt=2\sqrt t \).

Teacher's Note:
1. sin x cos x = tan x cos^2 x.
2. The numerator sec^2 x is dt.

Answer: (b) \( \frac1{\ln2} \)

Explanation:
1. Put \( t=2^x \), \( dt=2^x\ln2\,dx \).
2. The integral is \( \frac1{\ln2}\int\frac{dt}{\sqrt{1-t^2}}=\frac{\sin^{-1}t}{\ln2} \).
3. So \( K=\frac1{\ln2} \).

Teacher's Note:
1. 4^x = (2^x)^2.
2. dt contains the factor ln 2.

Answer: (a) \( \frac1{\sqrt2} \)

Explanation:
1. Put \( x=\tan\theta \): \( \int\cos\theta\,d\theta=\sin\theta=\frac{x}{\sqrt{1+x^2}} \).
2. With y(0) = 0 there is no constant. At x = 1: \( y=\frac{1}{\sqrt2} \).

Teacher's Note:
1. Trigonometric substitution x = tan theta.
2. The constant is zero because y(0) = 0.

Answer: (b) \( \frac16\sec^32x-\frac12\sec2x+C \)

Explanation:
1. Write \( \tan^32x\sec2x=(\sec^22x-1)\sec2x\tan2x \).
2. Put \( t=\sec2x \), \( dt=2\sec2x\tan2x\,dx \).
3. The integral is \( \frac12\int(t^2-1)dt=\frac{t^3}{6}-\frac t2 \).

Teacher's Note:
1. Keep one factor of sec tan for the differential.
2. Remember the factor 2 from the chain rule.

Answer: (b) \( \frac4{35},\ -\frac8{25},\ \frac4{15} \)

Explanation:
1. Put \( u=1+x^{5/2} \), so \( du=\frac52x^{3/2}dx \) and \( x^{13/2}=x^{3/2}(u-1)^2 \).
2. The integral is \( \frac25\int(u-1)^2u^{1/2}du=\frac25\int(u^{5/2}-2u^{3/2}+u^{1/2})du \).
3. This gives \( \frac4{35}u^{7/2}-\frac8{25}u^{5/2}+\frac4{15}u^{3/2} \).

Teacher's Note:
1. Write x^(13/2) as x^(3/2) times (x^(5/2))^2.
2. Expand (u - 1)^2 before integrating.

Answer: (d) \( \ln|x|-\frac27\ln|1+x^7|+C \)

Explanation:
1. Split: \( \frac{1-x^7}{x(1+x^7)}=\frac1x-\frac{2x^6}{1+x^7} \).
2. The second term integrates to \( \frac27\ln|1+x^7| \).
3. So the answer is \( \ln|x|-\frac27\ln|1+x^7|+C \).

Teacher's Note:
1. Write 1 - x^7 as 2 - (1 + x^7).
2. The numerator 2x^6 is nearly the derivative of 1 + x^7.

Answer: (b) \( -xe^{-x}+C \)

Explanation:
1. By parts with \( u=x-1 \) and \( dv=e^{-x}dx \): \( -(x-1)e^{-x}+\int e^{-x}dx \).
2. This equals \( -(x-1)e^{-x}-e^{-x}=-xe^{-x} \).

Teacher's Note:
1. Integrate by parts or differentiate the options.
2. d/dx(-x e^(-x)) = (x - 1) e^(-x).

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

Explanation:
1. Write the integrand as \( e^{\tan^{-1}x}\left(1+\frac{x}{1+x^2}\right) \).
2. Note \( \frac{d}{dx}\left(xe^{\tan^{-1}x}\right)=e^{\tan^{-1}x}\left(1+\frac{x}{1+x^2}\right) \).
3. Hence the integral is \( xe^{\tan^{-1}x}+C \).

Teacher's Note:
1. Use the form integral of (f + x f') with f = e^(arctan x).
2. Check by the product rule.

Answer: (c) \( f(x)g'(x)-f'(x)g(x)+C \)

Explanation:
1. Consider \( \frac{d}{dx}(fg'-f'g)=f'g'+fg''-f''g-f'g'=fg''-f''g \).
2. Therefore the integral is \( fg'-f'g+C \).

Teacher's Note:
1. Differentiate fg' - f'g and compare.
2. The middle terms f'g' cancel.

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

Explanation:
1. By parts with \( u=\ln x \), \( dv=\frac{x\,dx}{(x^2-1)^{3/2}} \), so \( v=-\frac1{\sqrt{x^2-1}} \).
2. Then \( -\frac{\ln x}{\sqrt{x^2-1}}+\int\frac{dx}{x\sqrt{x^2-1}} \).
3. The last integral is \( \sec^{-1}x \).

Teacher's Note:
1. Choose ln x as the first function.
2. The remaining integral is a standard inverse secant.

Answer: (b) \( -x\cos x+C \)

Explanation:
1. Since \( e^{\ln\sin x}=\sin x \), the integrand is \( x\sin x-\cos x \).
2. By parts \( \int x\sin x\,dx=-x\cos x+\sin x \).
3. Subtract \( \int\cos x\,dx=\sin x \) to get \( -x\cos x+C \).

Teacher's Note:
1. Simplify e^(ln sin x) first.
2. The sin x terms cancel.

Answer: (d) \( \frac2{\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3}+C \)

Explanation:
1. Complete the square: \( x^2+x+1=\left(x+\frac12\right)^2+\frac34 \).
2. Using \( \int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\frac ua \) with \( a=\frac{\sqrt3}2 \), the result is \( \frac2{\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3} \).

Teacher's Note:
1. Complete the square in the denominator.
2. 1/a with a = sqrt3/2 gives 2/sqrt3.

Answer: (c) \( -\left(1+\frac1{x^4}\right)^{1/4}+C \)

Explanation:
1. Write \( (x^4+1)^{3/4}=x^3(1+x^{-4})^{3/4} \), so the integrand is \( x^{-5}(1+x^{-4})^{-3/4} \).
2. Put \( t=1+x^{-4} \), \( dt=-4x^{-5}dx \): \( -\frac14\int t^{-3/4}dt=-t^{1/4} \).

Teacher's Note:
1. Factor x^4 out of the bracket.
2. The derivative of x^(-4) is -4 x^(-5).

Answer: (a) \( \frac13\ln\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C \)

Explanation:
1. Put \( t=\sqrt{1-x^3} \), so \( x^3=1-t^2 \) and \( 3x^2dx=-2t\,dt \).
2. Then \( \frac{dx}{x\sqrt{1-x^3}}=\frac{x^2dx}{x^3t}=\frac{-2\,dt}{3(1-t^2)} \).
3. Integrating gives \( \frac13\ln\left|\frac{t-1}{t+1}\right| \).

Teacher's Note:
1. Rationalise with t = sqrt(1 - x^3).
2. Multiply numerator and denominator by x^2.

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

Explanation:
1. Rationalise: \( \sqrt{\frac{e^x-1}{e^x+1}}=\frac{e^x-1}{\sqrt{e^{2x}-1}} \).
2. Put \( t=e^x \): \( \int\frac{t-1}{t\sqrt{t^2-1}}dt=\int\frac{dt}{\sqrt{t^2-1}}-\int\frac{dt}{t\sqrt{t^2-1}} \).
3. These are \( \ln(t+\sqrt{t^2-1})-\sec^{-1}t \).

Teacher's Note:
1. Multiply numerator and denominator by sqrt(e^x - 1).
2. Split the fraction after t = e^x.

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

Explanation:
1. Partial fractions: \( \frac1{x^3(x+1)}=\frac1{x^3}-\frac1{x^2}+\frac1x-\frac1{x+1} \).
2. Integrate: \( -\frac1{2x^2}+\frac1x+\ln\left|\frac x{x+1}\right| \).
3. So A = -1/2 and B = 1.

Teacher's Note:
1. Resolve into partial fractions.
2. The integral of 1/x^3 is -1/(2x^2).