JEE Mathematics Indefinite Integration MCQs Set 12

Answer: (d) \( 4\sqrt2 \)

Explanation:
1. \( 1+\sin t=\left(\sin\frac t2+\cos\frac t2\right)^2=2\sin^2\left(\frac t2+\frac\pi4\right) \), so with \( t=\frac x2 \) the integrand is \( \sqrt2\sin\left(\frac x4+\frac\pi4\right) \).
2. Integrate: \( -4\sqrt2\cos\left(\frac x4+\frac\pi4\right)=4\sqrt2\sin\left(\frac x4-\frac\pi4\right) \).

Teacher's Note:
1. Write 1 + sin t as a perfect square using half angles.
2. -cos(theta + pi/4) = sin(theta - pi/4).

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

Explanation:
1. Put \( x=\tan\theta \): \( \int\frac{\sec^2\theta\,d\theta}{\sec^3\theta}=\int\cos\theta\,d\theta=\sin\theta \).
2. \( y=\frac x{\sqrt{1+x^2}}+C \) and \( y(0)=0 \) gives \( C=0 \).
3. At \( x=1 \): \( y=\frac1{\sqrt2} \).

Teacher's Note:
1. The substitution x = tan(theta) handles powers of 1 + x^2.
2. Use the initial condition to fix the constant.

Answer: (c) \( \frac12\ln(e^{2x}+1)-\frac{\cot^{-1}(e^x)}{e^x}-x+c \)

Explanation:
1. Integrate by parts with \( u=\cot^{-1}(e^x) \) and \( dv=e^{-x}dx \): \( =-e^{-x}\cot^{-1}(e^x)-\int\frac{dx}{1+e^{2x}} \) (since \( \frac{d}{dx}\cot^{-1}e^x=-\frac{e^x}{1+e^{2x}} \)).
2. \( \int\frac{dx}{1+e^{2x}}=x-\frac12\ln(1+e^{2x}) \).
3. Result: \( \frac12\ln(e^{2x}+1)-\frac{\cot^{-1}(e^x)}{e^x}-x+c \).

Teacher's Note:
1. Integration by parts with the inverse trigonometric factor as u.
2. dx/(1 + e^(2x)) = dx - e^(2x) dx/(1 + e^(2x)).

Answer: (d) \( e^{\tan\theta}\cos\theta+c \)

Explanation:
1. Try \( \frac{d}{d\theta}\left(e^{\tan\theta}\cos\theta\right)=e^{\tan\theta}\left(\sec^2\theta\cos\theta-\sin\theta\right)=e^{\tan\theta}(\sec\theta-\sin\theta) \).
2. So the integral is \( e^{\tan\theta}\cos\theta+c \).

Teacher's Note:
1. Look for a product e^(tan) times a trigonometric function whose derivative matches.
2. sec^2 x cos x = sec x.

Answer: (c) \( \ln x-\frac27\ln(1+x^7)+c \)

Explanation:
1. \( \frac{1-x^7}{x(1+x^7)}=\frac1x-\frac{2x^6}{1+x^7} \) (since \( 1-x^7=(1+x^7)-2x^7 \)).
2. Integrate: \( \ln x-\frac27\ln(1+x^7)+c \).

Teacher's Note:
1. Split the numerator as (1 + x^7) - 2x^7.
2. The second integral has the form f'/f with f = 1 + x^7.

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

Explanation:
1. Put \( u=\sqrt{1+x^2} \), so \( x\,dx=u\,du \) and the denominator is \( \sqrt{u^2+u^3}=u\sqrt{1+u} \).
2. \( \int\frac{u\,du}{u\sqrt{1+u}}=\int\frac{du}{\sqrt{1+u}}=2\sqrt{1+u} \).
3. So the result is \( 2\sqrt{1+\sqrt{1+x^2}}+c \).

Teacher's Note:
1. Substitute u = sqrt(1 + x^2) so that x dx = u du.
2. Factor u out of the radical in the denominator.

Answer: (b) \( -\frac x{x^4+x+1}+c \)

Explanation:
1. Differentiate the candidate \( -\frac x{x^4+x+1} \): \( -\frac{(x^4+x+1)-x(4x^3+1)}{(x^4+x+1)^2}=-\frac{1-3x^4}{(x^4+x+1)^2}=\frac{3x^4-1}{(x^4+x+1)^2} \).

Teacher's Note:
1. Integrals of this type are of the form (u/v)' where the numerator is v - x v'.
2. Verify a candidate by differentiating.

Answer: (b) \( e^x+\sin x \)

Explanation:
1. Let \( h(x)=e^x+\sin x+x \), \( h'(x)=e^x+\cos x+1 \). The numerator is \( \cos x-\sin x+1-x=h'(x)-h(x) \).
2. So the integrand is \( \frac{h'}{h}-1 \) and the integral is \( \ln h(x)-x+C \).
3. Thus \( f=h \), \( g=-x \) and \( f+g=e^x+\sin x \).

Teacher's Note:
1. Check whether the numerator is the derivative of the denominator minus the denominator.
2. Identify f and g by comparing with ln f + g.

Answer: (b) \( 2\ln\sin\frac x2+c \)

Explanation:
1. \( 1+2\cot^2x+2\cot x\csc x=(\cot x+\csc x)^2 \) (using \( 1+\cot^2x=\csc^2x \)).
2. So the integral is \( \int(\cot x+\csc x)dx=\ln\sin x+\ln\tan\frac x2+c \).
3. This is \( \ln\left(2\sin^2\frac x2\right)+c=2\ln\sin\frac x2+c' \).

Teacher's Note:
1. Complete the square under the radical using cot^2 + 1 = csc^2.
2. The integral of csc x is ln tan(x/2).

Answer: (d) \( \frac{\cos3x}4+C \)

Explanation:
1. \( \sin^3\theta=\frac{3\sin\theta-\sin3\theta}4 \).
2. The three angles differ by \( \frac{2\pi}3 \), so \( \sum\sin\theta=0 \) and each \( \sin3\theta=\sin3x \) (since \( 3\cdot\frac{2\pi}3=2\pi \)).
3. \( f(x)=-\frac34\sin3x \), and its primitive is \( \frac{\cos3x}4+C \).

Teacher's Note:
1. Reduce the cubes with the triple-angle formula.
2. The three sines sum to zero because the angles are equally spaced.

Answer: (b) \( c-\frac{\sqrt[5]{1+x^5}}{x} \)

Explanation:
1. Write \( (x^5+1)^{4/5}=x^4(1+x^{-5})^{4/5} \) so the integrand is \( \frac{1}{x^6(1+x^{-5})^{4/5}} \).
2. Put \( t=1+x^{-5} \), \( dt=-5x^{-6}dx \): \( -\frac15\int t^{-4/5}dt=-t^{1/5} \).
3. So the result is \( -\left(1+x^{-5}\right)^{1/5}+c=c-\frac{\sqrt[5]{1+x^5}}{x} \).

Teacher's Note:
1. Factor x^5 out of the bracket so the substitution t = 1 + x^(-5) works.
2. x^(-5) inside the root becomes 1/x outside after multiplying by x.

Answer: (a) \( \sqrt{1+x^2}\,\ln\left(x+\sqrt{1+x^2}\right)-x+c \)

Explanation:
1. Integrate by parts with \( u=\ln\left(x+\sqrt{1+x^2}\right) \) and \( dv=\frac x{\sqrt{1+x^2}}dx \), so \( v=\sqrt{1+x^2} \) and \( du=\frac{dx}{\sqrt{1+x^2}} \).
2. Result: \( \sqrt{1+x^2}\ln\left(x+\sqrt{1+x^2}\right)-\int dx=\sqrt{1+x^2}\ln\left(x+\sqrt{1+x^2}\right)-x+c \).

Teacher's Note:
1. The derivative of ln(x + sqrt(1+x^2)) is 1/sqrt(1+x^2), which cancels nicely after parts.
2. Choose dv to be the algebraic part x/sqrt(1+x^2).

Answer: (c) \( \frac12\ln^2x-x+C \)

Explanation:
1. \( f(x)=\frac{\ln x-x}{x}=\frac{\ln x}x-1 \).
2. \( \int\frac{\ln x}xdx=\frac12\ln^2x \) and \( \int1\,dx=x \).
3. Primitive \( \frac12\ln^2x-x+C \).

Teacher's Note:
1. Simplify ln(x/e^x) = ln x - x first.
2. The integral of (ln x)/x is half of ln squared x.

Answer: (c) \( \frac{(x^2+1)^{\ln2+1}}{2(\ln2+1)}+C \)

Explanation:
1. \( 2^{\ln u}=u^{\ln2} \) with \( u=x^2+1 \), so the integrand is \( x(x^2+1)^{\ln2} \).
2. With \( u=x^2+1 \), \( x\,dx=\tfrac12du \): \( \frac12\int u^{\ln2}du=\frac{u^{\ln2+1}}{2(\ln2+1)} \).

Teacher's Note:
1. Use a^(ln u) = u^(ln a).
2. Then substitute u = x^2 + 1.

Answer: (b) \( -e\cos e^{x-1}+C \)

Explanation:
1. Put \( t=e^{x-1} \), so \( e^x=et \) and \( dt=e^{x-1}dx \), i.e. \( e^xdx=e\,dt \).
2. \( \int\sin t\cdot e\,dt=-e\cos t+C=-e\cos e^{x-1}+C \).

Teacher's Note:
1. Substitute t = e^(x-1) and note e^x = e t.
2. The factor e remains outside the integral.

Answer: (b) \( \frac{2f(x)g(x)}{1+(g(x))^2} \)

Explanation:
1. Since \( g'=f \): \( \frac{d}{dx}\ln(1+g^2)=\frac{2gg'}{1+g^2}=\frac{2f(x)g(x)}{1+(g(x))^2} \).

Teacher's Note:
1. Differentiate the given function using the chain rule and g' = f.
2. An antiderivative is any function whose derivative is the required integrand.

Answer: (c) \( \sin x-2(\sin x)^{-1}-6\tan^{-1}(\sin x)+c \)

Explanation:
1. Put \( t=\sin x \), \( \cos x\,dx=dt \): the integrand becomes \( \frac{(1-t^2)(2-t^2)}{t^2(1+t^2)}dt \) (since \( \cos^2x(1+\cos^2x)=(1-t^2)(2-t^2) \)).
2. \( \frac{t^4-3t^2+2}{t^2(t^2+1)}=1+\frac2{t^2}-\frac6{t^2+1} \).
3. Integrate: \( t-\frac2t-6\tan^{-1}t \), i.e. \( \sin x-\frac2{\sin x}-6\tan^{-1}(\sin x)+c \).

Teacher's Note:
1. Factor out one cos x and write the rest in terms of sin x.
2. Divide polynomial and use partial fractions.

Answer: (b) \( x\cdot\int\frac{dx}x=x\ln|x|+Cx \)

Explanation:
1. \( \int\frac{dx}x=\ln|x|+C \), and multiplying by x gives \( x\ln|x|+Cx \) (the constant gets multiplied too).
2. In (c) and (d): \( \frac1{\cos x}(\sin x+C)=\tan x+C\sec x \), not \( \tan x+C \) or \( x+C \).

Teacher's Note:
1. When an indefinite integral is multiplied by a function, the arbitrary constant is multiplied as well.
2. Check each option by actually multiplying out.

Answer: (a) \( \frac{2(x^2-1)^{5/4}}{45}(4+5x^2)+C \)

Explanation:
1. Put \( t=x^2-1 \), \( x\,dx=\tfrac{dt}2 \) and \( x^2=t+1 \): \( \frac12\int(t+1)t^{1/4}dt=\frac12\left[\frac{4t^{9/4}}9+\frac{4t^{5/4}}5\right] \).
2. Factor \( t^{5/4} \): \( \frac{2t^{5/4}}{45}(5t+9)=\frac{2(x^2-1)^{5/4}}{45}(5x^2+4) \).

Teacher's Note:
1. Substitute t = x^2 - 1 and write x^2 = t + 1.
2. Combine the two fractions by taking t^(5/4) as a common factor.

Answer: (a) \( \frac{\sin(100x)(\sin x)^{100}}{100}+C \)

Explanation:
1. Differentiate \( F=\frac{\sin(100x)\sin^{100}x}{100} \): \( F'=\cos(100x)\sin^{100}x+\sin(100x)\sin^{99}x\cos x=\sin^{99}x[\cos(100x)\sin x+\sin(100x)\cos x] \).
2. The bracket is \( \sin(101x) \), so \( F'=\sin(101x)\sin^{99}x \).

Teacher's Note:
1. Use the reverse of the product rule: try sin(100x) sin^100 x.
2. The bracket is the sine addition formula.

Answer: (b) \( \frac14\ln\left(\frac{\ln x-x}{\ln x+x}\right)-\frac12\tan^{-1}\left(\frac{\ln x}x\right)+C \)

Explanation:
1. Divide numerator and denominator by \( x^4 \) and put \( t=\frac{\ln x}x \), \( dt=\frac{1-\ln x}{x^2}dx \): the integral becomes \( \int\frac{dt}{t^4-1} \).
2. \( \frac1{t^4-1}=\frac12\left(\frac1{t^2-1}-\frac1{t^2+1}\right) \), so the integral is \( \frac14\ln\left|\frac{t-1}{t+1}\right|-\frac12\tan^{-1}t+C \).
3. Substitute back \( t=\frac{\ln x}x \) to get the second option.

Teacher's Note:
1. The numerator (1 - ln x)/x^2 is exactly the derivative of (ln x)/x.
2. Partial fractions of 1/(t^4 - 1) give a logarithm and an arctangent.

Answer: (c) \( -\frac{x^q}{x^{p+q}+1}+C \)

Explanation:
1. The denominator is \( (x^{p+q}+1)^2 \). Consider \( \frac{d}{dx}\frac{x^q}{x^{p+q}+1}=\frac{qx^{q-1}(x^{p+q}+1)-x^q(p+q)x^{p+q-1}}{(x^{p+q}+1)^2} \).
2. The numerator simplifies to \( qx^{q-1}-px^{p+2q-1} \), which is the negative of the given numerator.
3. So the integral is \( -\frac{x^q}{x^{p+q}+1}+C \).

Teacher's Note:
1. Recognise the denominator as a perfect square and guess a quotient of the form x^q/(x^(p+q)+1).
2. Compare the derivative of the guess with the integrand to fix the sign.

Answer: (c) \( C+2B=0 \)

Explanation:
1. Differentiate the right side and divide by \( e^{3x} \): \( 3Ax^3+(3A+3B)x^2+(2B+3C)x+(C+3D) \).
2. Match coefficients with \( x^3-2x^2+5 \): \( A=\tfrac13,\ B=-1,\ C=\tfrac23,\ D=\tfrac{13}9 \).
3. Check: \( C+3D=5 \) true; \( A+B+\tfrac23=0 \) true; \( C+2B=-\tfrac43\ne0 \) false; \( A+B+C=0 \) true. The incorrect statement is \( C+2B=0 \).

Teacher's Note:
1. Differentiate the proposed antiderivative and compare coefficients.
2. Evaluate all four statements with the values found.

Answer: (c) \( \frac{\sin2x}8\left[\cos^22x+1-2\cos2x\right]+C \)

Explanation:
1. Reduction formula: \( I_n=-\frac{\sin^{n-1}x\cos x}n+\frac{n-1}nI_{n-2} \). For \( n=6 \): \( 6I_6=-\sin^5x\cos x+5I_4 \).
2. So \( 5I_4-6I_6=\sin^5x\cos x+C \).
3. Now \( \sin^5x\cos x=\sin^4x\cdot\frac{\sin2x}2=\left(\frac{1-\cos2x}2\right)^2\frac{\sin2x}2=\frac{\sin2x}8(1-\cos2x)^2 \), which is option (c).

Teacher's Note:
1. Use the standard reduction formula for the integral of sin^n.
2. Convert sin^4 x with the double angle formula.

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

Explanation:
1. For \( x>0 \): \( \sec^{-1}\sqrt{1+x^2}=\tan^{-1}x \) and \( \cos^{-1}\frac{1-x^2}{1+x^2}=2\tan^{-1}x \).
2. Put \( t=\tan^{-1}x \): \( \int e^t(t^2+2t)dt=e^tt^2+C \) (form \( e^t(f+f') \)).
3. So the result is \( e^{\tan^{-1}x}(\tan^{-1}x)^2+C=e^{\tan^{-1}x}\left(\sec^{-1}\sqrt{1+x^2}\right)^2+C \).

Teacher's Note:
1. Simplify the inverse trigonometric functions for x > 0.
2. Use the e^t (f + f') pattern.

Answer: (b) \( \tan^{-1}\left(x+\frac1x\right) \)

Explanation:
1. Divide numerator and denominator by \( x^2 \): \( \frac{(1-\frac1{x^2})dx}{(x^2+3+\frac1{x^2})\tan^{-1}(x+\frac1x)} \).
2. Put \( u=x+\frac1x \), \( du=\left(1-\frac1{x^2}\right)dx \) and \( x^2+3+\tfrac1{x^2}=u^2+1 \).
3. \( \int\frac{du}{(1+u^2)\tan^{-1}u}=\ln\left|\tan^{-1}u\right|+C \), so \( f(x)=\tan^{-1}\left(x+\frac1x\right) \).

Teacher's Note:
1. Divide through by x^2 so that the numerator becomes du.
2. The derivative of arctan u is 1/(1 + u^2).

Answer: (a) \( e^x\tan x+c \)

Explanation:
1. \( \frac{2\sin^2x-1}{\cos x}=-\frac{\cos2x}{\cos x} \) and \( \frac{\cos x(2\sin x+1)}{1+\sin x} \); combining over \( \cos x(1+\sin x) \) gives \( f(x)=\tan x \).
2. Use \( \int e^x(f+f')dx=e^xf(x)+c \): the integral is \( e^x\tan x+c \).

Teacher's Note:
1. Simplify f(x) first; here it collapses to tan x.
2. Apply the e^x (f + f') rule.

Answer: (b) \( -\left(u+\frac{u^2}2+\cdots+\frac{u^9}9\right) \)

Explanation:
1. \( I_n+I_{n-2}=\int\cot^{n-2}x(\cot^2x+1)dx=-\frac{\cot^{n-1}x}{n-1} \).
2. The given sum equals \( \sum_{n=2}^{10}(I_n+I_{n-2}) \), because each of \( I_2,\dots,I_8 \) appears twice and \( I_0,I_1,I_9,I_{10} \) once.
3. Hence the sum is \( -\sum_{n=2}^{10}\frac{u^{n-1}}{n-1}=-\left(u+\frac{u^2}2+\cdots+\frac{u^9}9\right) \).

Teacher's Note:
1. Find the recurrence I_n + I_(n-2) first.
2. Check that the coefficients in the given sum telescope into pairs.

Answer: (b) B, C and D only

Explanation:
1. \( \frac{x^2+\cos^2x}{1+x^2}=1-\frac{\sin^2x}{1+x^2} \), so the integrand is \( \csc^2x-\frac1{1+x^2} \).
2. Integrate: \( -\cot x-\tan^{-1}x+c \).
3. (B) holds since \( \cot^{-1}x=\frac\pi2-\tan^{-1}x \) (constant absorbed); (C) holds since \( \frac{\csc x}{\sec x}=\cot x \); (D) holds since \( e^{\ln\tan^{-1}x}=\tan^{-1}x \). (A) has the wrong signs.

Teacher's Note:
1. Rewrite the numerator as (1 + x^2) - sin^2 x.
2. Constants can be absorbed, so cot^(-1) x and -tan^(-1) x differ by a constant.