JEE Mathematics Indefinite Integration MCQs Set 28

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

Explanation:
1. Write \(\sin x=\sin\left((x-\alpha)+\alpha\right)=\sin(x-\alpha)\cos\alpha+\cos(x-\alpha)\sin\alpha\).
2. The integrand becomes \(\cos\alpha+\sin\alpha\cot(x-\alpha)\), so the integral is \(x\cos\alpha+\sin\alpha\log\sin(x-\alpha)\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (c) \(\dfrac1{1-x^2}+c\)

Explanation:
1. The sum is \(\dfrac d{dx}\left(x^2+x^4+\dots\right)=\dfrac d{dx}\dfrac{x^2}{1-x^2}=\dfrac{2x}{(1-x^2)^2}\).
2. \(\displaystyle\int\dfrac{2x}{(1-x^2)^2}dx=\dfrac1{1-x^2}+c\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (d) \(\dfrac12\left(\log\dfrac{g(x)}{f(x)}\right)^2+C\)

Explanation:
1. Let \(u=\log\dfrac{g}{f}\); then \(du=\left(\dfrac{g'}g-\dfrac{f'}f\right)dx=\dfrac{fg'-f'g}{fg}dx\).
2. The integral is \(\displaystyle\int u\,du=\dfrac{u^2}2+C\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (a) \(e^x\sqrt{\dfrac{1+x^n}{1-x^n}}+C\)

Explanation:
1. Use \(\int e^x\left(g+g'\right)dx=e^xg\) with \(g=\sqrt{\dfrac{1+x^n}{1-x^n}}\).
2. Here \(g'=\dfrac{nx^{n-1}}{(1-x^n)\sqrt{1-x^{2n}}}\) and \(g+g\) simplifies to the given integrand, so the integral is \(e^xg+C\) (confirmed by differentiating the result).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (c) \(\dfrac{x^2}2\ln x-\dfrac{x^2}4+C\)

Explanation:
1. \(x^{1/n}-x^{1/(n+1)}\approx\ln x\left(\dfrac1n-\dfrac1{n+1}\right)=\dfrac{\ln x}{n(n+1)}\), so \(f(x)=\ln x\).
2. \(\displaystyle\int x\ln x\,dx=\dfrac{x^2}2\ln x-\dfrac{x^2}4+C\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

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

Explanation:
1. For \(0<x<1\), \(x^n\to0\) and \(x^{-n}\to\infty\), so \(f(x)=-1\).
2. \(-\displaystyle\int\sin^{-1}x\,dx=-\left[x\sin^{-1}x+\sqrt{1-x^2}\right]+C\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (d) \(\ln(\sec2\theta)\)

Explanation:
1. The logarithm is \(\ln\tan\left(\theta+\dfrac\pi4\right)\) with derivative \(\dfrac2{\cos2\theta}\).
2. By parts: \(\dfrac{\sin2\theta}2\ln(\cdot)-\displaystyle\int\tan2\theta\,d\theta=\dfrac{\sin2\theta}2\ln(\cdot)-\dfrac12\ln\sec2\theta+c\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (a) \(\dfrac2{\alpha-\beta}\sqrt{\dfrac{\beta-x}{x-\alpha}}+C\)

Explanation:
1. Put \(x-\alpha=\dfrac1t\); the integral becomes a multiple of \(\displaystyle\int\dfrac{dt}{\sqrt{(\beta-\alpha)t-1}}\).
2. This gives \(\dfrac2{\alpha-\beta}\sqrt{\dfrac{\beta-x}{x-\alpha}}+C\) (verified by differentiation).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (b) \(\sqrt{1+\tan x+\cot x}\)

Explanation:
1. Put \(t=\tan x\); the integrand becomes \(\dfrac{(1-t)\,dt}{(1+t)\sqrt{t^3+t^2+t}}\) after \(\tan(\frac\pi4-x)=\dfrac{1-t}{1+t}\).
2. Rewriting \(\dfrac{1-t}{(1+t)\sqrt{t}\sqrt{t^2+t+1}}\) with \(v=\sqrt{t+\frac1t+1}\) gives \(-2\tan^{-1}v\) (checked by differentiation).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (c) \((\sin x+\cos x)^{n-1}(\sin x-\cos x)\)

Explanation:
1. Write \((s+c)^n=(s+c)^{n-1}(s+c)\) and integrate by parts with \(\int(s+c)dx=s-c\).
2. Using \((s-c)^2=2-(s+c)^2\) one gets \(nI_n=(s+c)^{n-1}(s-c)+2(n-1)I_{n-2}\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (d) \(\dfrac{(\sin x-\cos x)}{15}\left[3t^4+8t^2+32\right]+C\)

Explanation:
1. \(I_3=\dfrac{t^2(s-c)+4(s-c)}3=\dfrac{(s-c)(t^2+4)}3\).
2. \(I_5=\dfrac{t^4(s-c)+8I_3}5=\dfrac{(s-c)}{15}\left(3t^4+8t^2+32\right)\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (a) infinite

Explanation:
1. \(x^2-1>0\) gives \(S=(-\infty,-1)\cup(1,\infty)\).
2. On \((1,\infty)\) the condition \(f(2)=0\) fixes the constant, but on \((-\infty,-1)\) the additive constant is arbitrary, giving infinitely many functions.

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (d) \(1\)

Explanation:
1. \(\dfrac d{dx}\left[\sin100x\,\sin^{100}x\right]=100\sin^{99}x\left(\cos100x\sin x+\sin100x\cos x\right)=100\sin^{99}x\sin101x\).
2. So \(\lambda=100\), \(\mu=100\) and \(\dfrac\lambda\mu=1\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (c) \(\dfrac1{2\sqrt3}\tan^{-1}\left(\dfrac{\sqrt3}2\tan x\right)+c\)

Explanation:
1. Divide by \(\cos^2x\): \(\displaystyle\int\dfrac{\sec^2x\,dx}{3\tan^2x+4}\).
2. With \(t=\tan x\): \(\displaystyle\int\dfrac{dt}{3t^2+4}=\dfrac1{2\sqrt3}\tan^{-1}\dfrac{\sqrt3t}2+c\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (a) \(\dfrac12x|x|\ln|x|-\dfrac14x|x|+c\)

Explanation:
1. For \(x>0\) the integral is \(\dfrac{x^2}2\ln x-\dfrac{x^2}4\).
2. For \(x<0\) the integrand is \(-x\ln(-x)\) and the antiderivative is \(-\dfrac{x^2}2\ln|x|+\dfrac{x^2}4\).
3. Both cases are given by \(\dfrac12x|x|\ln|x|-\dfrac14x|x|\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (b) is greater than \(2\)

Explanation:
1. The integrand is \(\dfrac{x^{5/2}}{x^{7/2}(1+x^{5/2})}=\dfrac1{x(1+x^{5/2})}\).
2. With \(u=x^{5/2}\), \(\dfrac{dx}x=\dfrac2{5}\dfrac{du}u\) and the integral is \(\dfrac25\ln\dfrac{u}{1+u}\): \(\lambda=\dfrac25\), \(a=\dfrac52\), so \(a+\lambda=2.9>2\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

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

Explanation:
1. \(A^n=2^{n-1}x^n\begin{bmatrix}1&1\\1&1\end{bmatrix}\), so \(e^A=I+\dfrac{e^{2x}-1}2\begin{bmatrix}1&1\\1&1\end{bmatrix}\).
2. Hence \(f=e^{2x}+1\), \(g=e^{2x}-1\) and \(\dfrac gf=\tanh x\), whose integral is \(\log\cosh x=\log\left(e^x+e^{-x}\right)+c'\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.

Answer: (c) \(\dfrac{e^{2x}}5\left(2\sin x-\cos x\right)+c\)

Explanation:
1. \(g+1=e^{2x}\), so we need \(\displaystyle\int e^{2x}\sin x\,dx\).
2. By the standard formula \(\displaystyle\int e^{ax}\sin x\,dx=\dfrac{e^{ax}(a\sin x-\cos x)}{a^2+1}\) with \(a=2\).

Teacher's Note:
1. Differentiate the proposed answer to confirm it reproduces the integrand.
2. Look for a substitution or an integration by parts that matches the structure of the integrand.