JEE Mathematics Indefinite Integration MCQs Set 09

Answer: (c) \( \dfrac12\{\phi(x)+e^xf(x)\} \)

Explanation:
1. \( e^xf=\frac12\left[e^x(f+f')+e^x(f-f')\right] \).
2. \( \int e^x(f+f')dx=e^xf \) and \( \int e^x(f-f')dx=\phi \), so \( \int e^xf\,dx=\frac12\{\phi(x)+e^xf(x)\} \).

Teacher's Note:
1. Write e to the x f as half of a sum and a difference.
2. Use the f plus f dash rule for one of the two parts.

Answer: (b) \( xf^{-1}(x)-g(f^{-1}(x)) \)

Explanation:
1. Put \( x=f(t) \), so \( dx=f'(t)dt \) and \( \int f^{-1}(x)dx=\int t f'(t)dt \).
2. By parts, \( tf(t)-\int f(t)dt=tf(t)-g(t) \). Substituting \( t=f^{-1}(x) \) gives \( xf^{-1}(x)-g(f^{-1}(x))+c \).

Teacher's Note:
1. Substitute x = f(t) to remove the inverse function.
2. Then integrate t f dash t by parts.

Answer: (a) \( 2 \sqrt{x} \sin{\left(\sqrt{x} \right)} + 2 \cos{\left(\sqrt{x} \right)}+c \)

Explanation:
1. Put \( \sqrt x=t \): \( 2\int t\cos t\,dt=2(t\sin t+\cos t) \).
2. So the result is \( 2\left[\sqrt x\sin\sqrt x+\cos\sqrt x\right]+c \).

Teacher's Note:
1. Same method as the sine case.
2. Check the sign of the cosine term.

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

Explanation:
1. By parts with \( dv=\frac{x}{\sqrt{1-x^2}}dx \), \( v=-\sqrt{1-x^2} \).
2. \( -\sqrt{1-x^2}\sin^{-1}x+\int\sqrt{1-x^2}\cdot\frac{1}{\sqrt{1-x^2}}dx=x-\sqrt{1-x^2}\sin^{-1}x+c \).

Teacher's Note:
1. Choose dv so that it integrates easily.
2. The root cancels in the second integral.

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

Explanation:
1. By parts with \( v=-\frac{1}{\sqrt{1+x^2}} \): \( -\frac{\tan^{-1}x}{\sqrt{1+x^2}}+\int\frac{dx}{(1+x^2)^{3/2}} \).
2. \( \int\frac{dx}{(1+x^2)^{3/2}}=\frac{x}{\sqrt{1+x^2}} \), so the result is \( \frac{x-\tan^{-1}x}{\sqrt{1+x^2}}+c \).

Teacher's Note:
1. Integrate x over (1 + x squared) to the 3/2 first.
2. The last integral follows from x = tan theta.

Answer: (d) \( x \sin{\left(\log{\left(x \right)} \right)}+c \)

Explanation:
1. \( \frac{d}{dx}\left[x\sin(\log x)\right]=\sin(\log x)+x\cos(\log x)\cdot\frac1x=\sin(\log x)+\cos(\log x) \).
2. So the integral is \( x\sin(\log x)+c \).

Teacher's Note:
1. Try x times a trigonometric function of log x.
2. Differentiate to confirm.

Answer: (d) \( \dfrac12\sin2\theta\log\tan\left(\dfrac\pi4+\theta\right)-\dfrac12\log\sec2\theta \)

Explanation:
1. By parts: \( \frac12\sin2\theta\log\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)-\frac12\int\sin2\theta\cdot\frac{d}{d\theta}\log\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta \).
2. The derivative of the logarithm is \( \frac{2}{\cos2\theta} \), so the second integral is \( -\int\tan2\theta\,d\theta=-\frac12\log\sec2\theta \).
3. Since \( \frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\tan\left(\frac\pi4+\theta\right) \), the result is \( \frac12\sin2\theta\log\tan\left(\frac\pi4+\theta\right)-\frac12\log\sec2\theta+c \).

Teacher's Note:
1. Take the logarithm as u and cos 2 theta as dv.
2. The derivative of the log simplifies to 2 over cos 2 theta.

Answer: (b) \( x \log{\left(\log{\left(x \right)} \right)} - \frac{x}{\log{\left(x \right)}}+c \)

Explanation:
1. By parts, \( \int\log(\log x)dx=x\log(\log x)-\int\frac{dx}{\log x} \).
2. Also \( \frac{d}{dx}\left(\frac{x}{\log x}\right)=\frac{1}{\log x}-\frac{1}{(\log x)^2} \), so \( \int\left[\frac1{\log x}-\frac{1}{(\log x)^2}\right]dx=\frac{x}{\log x} \).
3. Hence the integral is \( x\log(\log x)-\frac{x}{\log x}+c \).

Teacher's Note:
1. The terms 1 over log x cancel between the two parts.
2. Use the derivative of x over log x.

Answer: (a) \( 1 \)

Explanation:
1. \( g'(x)=\frac{2\sin x-\sin2x}{x^3} \).
2. Using \( \sin x=x-\frac{x^3}{6}+\dots \) and \( \sin2x=2x-\frac{8x^3}{6}+\dots \), the numerator is \( \frac{-2+8}{6}x^3+\dots=x^3+\dots \).
3. Hence \( \lim_{x\to0}g'(x)=1 \).

Teacher's Note:
1. Differentiating an integral returns the integrand.
2. Use series expansions to evaluate the limit.

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

Explanation:
1. Put \( u=\cos^2x \), \( du=-\sin2x\,dx \) and \( \log\cos x=\frac12\log u \): \( -\frac12\int\log u\,du \).
2. \( -\frac12(u\log u-u)=-\cos^2x\log\cos x+\frac12\cos^2x=\cos^2x\left(\frac12-\log\cos x\right)+k \).

Teacher's Note:
1. Notice sin 2x dx is minus d of cos squared x.
2. Then integrate log u.

Answer: (b) \( \phi(x)=\dfrac{1+x^2}{2} \) and \( \psi(x)=-\dfrac{1+x^2}{2} \)

Explanation:
1. Put \( u=1+x^2 \), \( x\,dx=\frac{du}{2} \): \( \frac12\int\log u\,du=\frac12(u\log u-u) \).
2. So \( \int x\log(1+x^2)dx=\frac{1+x^2}{2}\log(1+x^2)-\frac{1+x^2}{2}+c \), giving \( \phi=\frac{1+x^2}{2} \) and \( \psi=-\frac{1+x^2}{2} \).

Teacher's Note:
1. Substitute u = 1 + x squared and integrate log u.
2. The functions phi and psi are read off the result.

Answer: (a) \( f(x)=\tan^{-1}x,\ A=-1 \)

Explanation:
1. By parts with \( v=\sqrt{1+x^2} \): \( \sqrt{1+x^2}\tan^{-1}x-\int\sqrt{1+x^2}\cdot\frac{1}{1+x^2}dx \).
2. \( \int\frac{dx}{\sqrt{1+x^2}}=\log\left(x+\sqrt{x^2+1}\right) \), so \( f(x)=\tan^{-1}x \) and \( A=-1 \).

Teacher's Note:
1. Integrate x over the root first.
2. The remaining integral is a standard logarithmic form.

Answer: (d) None of these

Explanation:
1. \( \int x\log(x+1)dx=\frac{x^2-1}{2}\log(x+1)-\frac{x^2}{4}+\frac x2 \) and \( \int x\log x\,dx=\frac{x^2}{2}\log x-\frac{x^2}{4} \).
2. Subtracting, the integral is \( \frac{x^2-1}{2}\log(x+1)-\frac12x^2\log x+\frac12x+c \).
3. So \( f(x)=\frac{x^2-1}{2} \), \( g(x)=-\frac12\log x \) and \( A=\frac12 \); none of (a), (b), (c) holds.

Teacher's Note:
1. Split the logarithm as log (x + 1) minus log x.
2. Compare the coefficients with the given form.

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

Explanation:
1. By parts with \( v=\int\frac{dx}{(1-x^2)^{3/2}}=\frac{x}{\sqrt{1-x^2}} \): \( \frac{x\sin^{-1}x}{\sqrt{1-x^2}}-\int\frac{x}{1-x^2}dx \).
2. \( \int\frac{x}{1-x^2}dx=-\frac12\log(1-x^2) \), so the result is \( \frac{x}{\sqrt{1-x^2}}\sin^{-1}x+\frac12\log(1-x^2)+c \).

Teacher's Note:
1. Use x = sin theta to integrate 1 over (1 - x squared) to the 3/2.
2. The remaining integral is of the form f dash over f.

Answer: (a) \( \dfrac12\left[x^2F(x^2)-\int F(x^2)\,d(x^2)\right] \)

Explanation:
1. Put \( t=x^2 \), \( x\,dx=\frac{dt}{2} \): \( \int x^3f(x^2)dx=\frac12\int tf(t)\,dt \).
2. By parts, \( \frac12\left[tF(t)-\int F(t)dt\right]=\frac12\left[x^2F(x^2)-\int F(x^2)d(x^2)\right] \).

Teacher's Note:
1. Substitute t = x squared to reduce the power.
2. Then apply by parts with f as the second function.

Answer: (b) \( \frac{\tan^{-1}{\left(\frac{x}{3} + \frac{2}{3} \right)}}{3}+c \)

Explanation:
1. Complete the square: \( x^2+4x+13=(x+2)^2+3^2 \).
2. \( \int\frac{dx}{(x+2)^2+3^2}=\frac13\tan^{-1}\frac{x+2}{3}+c \).

Teacher's Note:
1. Complete the square in the denominator.
2. Use the standard arctangent integral.

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

Explanation:
1. \( x^2+8x+20=(x+4)^2+2^2 \).
2. \( \int\frac{dx}{(x+4)^2+2^2}=\frac12\tan^{-1}\frac{x+4}{2}+c \).

Teacher's Note:
1. Complete the square first.
2. Remember the factor 1 over a in the standard formula.

Answer: (a) \( \frac{x^{3}}{3} + \frac{x^{2}}{2} + x+c \)

Explanation:
1. \( x^4+x^2+1=(x^2+x+1)(x^2-x+1) \), so the integrand is \( x^2+x+1 \).
2. \( \int(x^2+x+1)dx=\frac{x^3}{3}+\frac{x^2}{2}+x+c \).

Teacher's Note:
1. Factorise the numerator as a difference of squares.
2. Cancel the common factor.

Answer: (a) \( \frac{\sqrt{5} \log{\left(\frac{\log{\left(x \right)} - \sqrt{5} + 2}{\log{\left(x \right)} + 2 + \sqrt{5}} \right)}}{10}+c \)

Explanation:
1. Put \( t=\log x \), \( dt=\frac{dx}{x} \): \( \int\frac{dt}{t^2+4t-1}=\int\frac{dt}{(t+2)^2-5} \).
2. \( =\frac{1}{2\sqrt5}\log\frac{t+2-\sqrt5}{t+2+\sqrt5}+c \).

Teacher's Note:
1. Substitute t = log x to remove x.
2. Use the formula for 1 over t squared minus a squared.

Answer: (a) \( \frac{\sqrt{6} \tan^{-1}{\left(\frac{\sqrt{6} \tan{\left(\frac{x}{2} \right)}}{6} \right)}}{6}+c \)

Explanation:
1. Put \( t=\tan\frac x2 \): \( \cos x=\frac{1-t^2}{1+t^2} \), \( dx=\frac{2dt}{1+t^2} \).
2. \( \int\frac{2dt}{7(1+t^2)+5(1-t^2)}=\int\frac{dt}{t^2+6}=\frac1{\sqrt6}\tan^{-1}\frac{t}{\sqrt6} \).

Teacher's Note:
1. Use the half angle substitution for a cosine denominator.
2. The denominator reduces to t squared plus 6.

Answer: (b) \( \frac{\log{\left(\tan{\left(\frac{x}{2} + \frac{\pi}{6} \right)} \right)}}{2}+c \)

Explanation:
1. \( \sin x+\sqrt3\cos x=2\sin\left(x+\frac\pi3\right) \).
2. \( \frac12\int\operatorname{cosec}\left(x+\frac\pi3\right)dx=\frac12\log\tan\left(\frac x2+\frac\pi6\right)+c \).

Teacher's Note:
1. Write the denominator as R sin (x + alpha).
2. The integral of cosec is log tan of half the angle.

Answer: (d) \( - \frac{\sqrt{2} \cot{\left(\frac{x}{2} + \frac{\pi}{8} \right)}}{2}+c \)

Explanation:
1. \( \sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right) \), so the denominator is \( \sqrt2\left[1+\sin\left(x-\frac\pi4\right)\right] \).
2. Using \( \int\frac{d\theta}{1+\sin\theta}=-\cot\left(\frac\theta2+\frac\pi4\right) \) with \( \theta=x-\frac\pi4 \), the result is \( -\frac1{\sqrt2}\cot\left(\frac x2+\frac\pi8\right)+c \).

Teacher's Note:
1. Write sin x minus cos x as a single sine.
2. Then use 1 plus sin theta equals 2 sin squared of a half angle.

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

Explanation:
1. Divide by \( \cos^2x \): \( \frac{\sec^2x}{\sec^2x+\tan^2x}=\frac{\sec^2x}{1+2\tan^2x} \).
2. Put \( t=\tan x \): \( \int\frac{dt}{1+2t^2}=\frac1{\sqrt2}\tan^{-1}(\sqrt2t) \).

Teacher's Note:
1. Divide numerator and denominator by cos squared x.
2. Then put t = tan x.

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

Explanation:
1. Divide by \( \cos^2x \): \( \frac{\sec^2x}{\sec^2x+1}=\frac{\sec^2x}{\tan^2x+2} \).
2. Put \( t=\tan x \): \( \int\frac{dt}{t^2+2}=\frac1{\sqrt2}\tan^{-1}\frac{t}{\sqrt2}+c \).

Teacher's Note:
1. Use the same method as for 1 over 1 + sin squared x.
2. Divide numerator and denominator by cos squared x.

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

Explanation:
1. Divide by \( \cos^2x \): \( \frac{\sec^2x}{(a\tan x+b)^2} \).
2. Put \( t=a\tan x+b \), \( dt=a\sec^2x\,dx \): \( \frac1a\int t^{-2}dt=-\frac{1}{a(a\tan x+b)}+c \).

Teacher's Note:
1. Divide numerator and denominator by cos squared x.
2. The factor a in dt gives the 1 over a.