JEE Mathematics Indefinite Integration MCQs Set 07

Answer: (a) \( \frac{\sqrt{1 - x^{4}}}{2} + \frac{\sin^{-1}{\left(x^{2} \right)}}{2}+c \)

Explanation:
1. Put \( t=x^2 \), \( x\,dx=\frac{dt}{2} \): \( \frac12\int\sqrt{\frac{1-t}{1+t}}\,dt=\frac12\int\frac{1-t}{\sqrt{1-t^2}}dt \).
2. \( \frac12\left[\sin^{-1}t+\sqrt{1-t^2}\right]=\frac12\left[\sin^{-1}x^2+\sqrt{1-x^4}\right]+c \).

Teacher's Note:
1. Rationalise by multiplying by 1 - t in numerator and denominator.
2. The second part uses a substitution in t squared.

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

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

Teacher's Note:
1. The factor 1 over cos squared is the derivative of tan x.
2. The derivative of 1 - t is minus 1, which cancels the minus from the power rule.

Answer: (b) \( \dfrac{\sec^px}{p}+c \)

Explanation:
1. Write \( \sec^px\tan x=\sec^{p-1}x\cdot\sec x\tan x \) and put \( t=\sec x \).
2. \( \int t^{p-1}dt=\frac{t^p}{p}=\frac{\sec^px}{p}+c \).

Teacher's Note:
1. Separate one factor sec x tan x, which is the derivative of sec x.
2. The remaining power is p minus 1.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. With \( x=a\tan\theta \), \( dx=a\sec^2\theta\,d\theta \) and \( x^2+a^2=a^2\sec^2\theta \), so the integral becomes \( \frac1a\int d\theta=\frac1a\tan^{-1}\frac xa \). The assertion is true.
2. Many integrals are done by parts, by partial fractions or by direct formulae, so the reason is false.

Teacher's Note:
1. Substituting x = a tan theta removes the sum x squared plus a squared.
2. Substitution is only one of several integration methods.

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

Explanation:
1. Put \( x=a\sin^2\theta \), \( dx=2a\sin\theta\cos\theta\,d\theta \); the integrand becomes \( \frac{\cos\theta}{\sin\theta} \).
2. \( 2a\int\cos^2\theta\,d\theta=a(\theta+\sin\theta\cos\theta) \), and \( \theta=\sin^{-1}\sqrt{x/a} \).
3. So the result is \( a\left[\sin^{-1}\sqrt{\frac xa}+\sqrt{\frac xa}\sqrt{\frac{a-x}{a}}\right]+C \).

Teacher's Note:
1. Use x = a sin squared theta for expressions with (a - x) over x.
2. Reduce cos squared with the double angle formula.

Answer: (b) \( \dfrac12\left[\log(\sin x-\cos x)+x\right]+c \)

Explanation:
1. Write \( \sin x=\frac12\left[(\sin x-\cos x)+(\sin x+\cos x)\right] \).
2. The integrand is \( \frac12+\frac12\cdot\frac{\sin x+\cos x}{\sin x-\cos x} \), and the second fraction has the derivative of the denominator in the numerator (with a minus sign): \( \frac{d}{dx}(\sin x-\cos x)=\cos x+\sin x \).
3. The integral is \( \frac12\left[x+\log(\sin x-\cos x)\right]+c \).

Teacher's Note:
1. Express sin x as half the sum of the denominator and its derivative.
2. This is the standard method for a sin x + b cos x over c sin x + d cos x.

Answer: (a) \( - \frac{x \sqrt{1 - x^{2}}}{2} + \frac{3 \sin^{-1}{\left(x \right)}}{2}+c \)

Explanation:
1. Write \( 1+x^2=2-(1-x^2) \): the integrand is \( \frac{2}{\sqrt{1-x^2}}-\sqrt{1-x^2} \).
2. \( 2\sin^{-1}x-\frac12\left[x\sqrt{1-x^2}+\sin^{-1}x\right]=\frac32\sin^{-1}x-\frac12x\sqrt{1-x^2}+c \).

Teacher's Note:
1. Rewrite the numerator in terms of the denominator's radicand.
2. Use the standard integral of root of 1 - x squared.

Answer: (b) \( K=\dfrac13,\ L=2,\ M=-1 \)

Explanation:
1. Put \( t=\tan x \): \( \sec^4x\operatorname{cosec}^2x\,dx=\frac{(1+t^2)^2}{t^2}dt=\left(t^{-2}+2+t^2\right)dt \).
2. Integrating: \( -\frac1t+2t+\frac{t^3}{3}=\frac13\tan^3x+2\tan x-\cot x \), so \( K=\frac13,\ L=2,\ M=-1 \).

Teacher's Note:
1. Convert everything to tan x using sec squared = 1 + tan squared.
2. Compare coefficients of tan cubed, tan and cot.

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

Explanation:
1. Put \( u=\log\frac{x+1}{x} \); then \( du=\left(\frac{1}{x+1}-\frac1x\right)dx=-\frac{dx}{x(x+1)} \).
2. \( -\int u\,du=-\frac{u^2}{2}=-\frac12\left[\log\frac{x+1}{x}\right]^2+c \).

Teacher's Note:
1. Combine the two logs into one and find its derivative.
2. The derivative produces the denominator factor with a minus sign.

Answer: (d) All of (a), (b) and (c)

Explanation:
1. \( \int\frac{dx}{\sqrt{1+\sin x}}=\frac{1}{\sqrt2}\int\operatorname{cosec}\left(\frac x2+\frac\pi4\right)dx=\sqrt2\log\tan\left(\frac x4+\frac\pi8\right) \), which is option (a).
2. Since \( \tan\frac\theta2=\operatorname{cosec}\theta-\cot\theta \), option (b) equals (a).
3. Since \( \sec\phi+\tan\phi=\tan\left(\frac\pi4+\frac\phi2\right) \) with \( \phi=\frac x2-\frac\pi4 \), option (c) equals (a) as well. So all three are correct.

Teacher's Note:
1. The same antiderivative can be written in several equivalent log forms.
2. Check equivalence with half-angle identities.

Answer: (b) \( \dfrac{1}{1+xe^x} \)

Explanation:
1. Put \( t=xe^x \); then \( dt=(x+1)e^xdx \) and \( \frac{(x+1)dx}{x}=\frac{dt}{xe^x}=\frac{dt}{t} \).
2. \( \int\frac{dt}{t(1+t)^2}=\int\left[\frac1t-\frac{1}{1+t}-\frac{1}{(1+t)^2}\right]dt=\log\frac{t}{1+t}+\frac{1}{1+t}=\log\left|1-\frac{1}{1+t}\right|+\frac{1}{1+t} \).
3. So \( f(x)=\frac{1}{1+xe^x} \).

Teacher's Note:
1. Substitute t = x e to the x and use partial fractions.
2. Rewrite log of t over 1 + t as log of 1 minus 1 over 1 + t.

Answer: (a) \( l^{r+1}(x)+c \)

Explanation:
1. \( \frac{d}{dx}l^{r+1}(x)=\frac{d}{dx}\log\left(l^r(x)\right)=\frac{(l^r)'}{l^r} \).
2. Applying this repeatedly, \( \frac{d}{dx}l^{r+1}(x)=\frac{1}{x\,l(x)\,l^2(x)\cdots l^r(x)} \), so the integral is \( l^{r+1}(x)+c \).

Teacher's Note:
1. Each extra log in the chain contributes one factor in the denominator.
2. Verify with r = 1: the integral of 1 over x log x is log log x.

Answer: (b) \( \log\sec\left(\dfrac{x^2-1}{2}\right) \)

Explanation:
1. With \( u=x^2-1 \): \( \frac{2\sin u-\sin2u}{2\sin u+\sin2u}=\frac{2\sin u(1-\cos u)}{2\sin u(1+\cos u)}=\tan^2\frac u2 \).
2. The integral is \( \int\tan\frac u2\cdot\frac{du}{2}=-\log\left|\cos\frac u2\right|=\log\sec\frac{x^2-1}{2}+c \).

Teacher's Note:
1. Factor 2 sin u from the numerator and the denominator.
2. Use x dx = du/2.

Answer: (b) \( \dfrac12x^2+k \)

Explanation:
1. Let \( \tan\theta=\sqrt{\frac{1-x}{1+x}} \). Then \( \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{(1+x)-(1-x)}{(1+x)+(1-x)}=x \).
2. So the integral is \( \int x\,dx=\frac{x^2}{2}+k \).

Teacher's Note:
1. Use cos 2 theta in terms of tan theta.
2. The inverse trigonometric expression simplifies to x itself.

Answer: (b) \( \sqrt[3]{\log_e(3y-2)} \)

Explanation:
1. \( y=\frac13e^{x^3}+C \). Using \( (0,1) \): \( 1=\frac13+C \), so \( C=\frac23 \).
2. \( 3y-2=e^{x^3} \), so \( x^3=\log_e(3y-2) \) and \( x=\sqrt[3]{\log_e(3y-2)} \).

Teacher's Note:
1. Use the given point to find the constant of integration.
2. Then solve the equation for x.

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

Explanation:
1. \( \cos x-\cos^3x=\cos x\sin^2x \), so the integrand is \( \frac{\sin x\sqrt{\cos x}}{\sqrt{1-\cos^3x}} \) for \( \sin x>0 \).
2. Put \( t=\cos^{3/2}x \); then \( dt=-\frac32\sqrt{\cos x}\sin x\,dx \), giving \( -\frac23\int\frac{dt}{\sqrt{1-t^2}}=\frac23\cos^{-1}t+c \).

Teacher's Note:
1. Factor the numerator and look for the derivative of cos to the 3/2.
2. The minus sign turns arcsine into arccosine.

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

Explanation:
1. For \( x>0 \), \( \sqrt{c^2x^2-(ax^2+b)^2}=x\sqrt{c^2-\left(ax+\frac bx\right)^2} \).
2. Put \( t=ax+\frac bx \); then \( dt=\left(a-\frac{b}{x^2}\right)dx=\frac{ax^2-b}{x^2}dx \), and the integral is \( \int\frac{dt}{\sqrt{c^2-t^2}}=\sin^{-1}\frac tc+k \).

Teacher's Note:
1. Divide out x from the root to expose the combination a x + b/x.
2. Its derivative matches the numerator over x squared.

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

Explanation:
1. Put \( \sqrt x=\cos\theta \), \( dx=-2\cos\theta\sin\theta\,d\theta \); then \( \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\tan\frac\theta2 \).
2. \( \tan\frac\theta2\sin\theta=1-\cos\theta \), so the integral is \( -2\int(1-\cos\theta)\cos\theta\,d\theta=-2\sin\theta+\theta+\sin\theta\cos\theta \).
3. Back-substituting: \( \cos^{-1}\sqrt x+\sqrt{1-x}\left(\sqrt x-2\right)+c \).

Teacher's Note:
1. Use x = cos squared theta for expressions with root x.
2. Half-angle tan reduces the square root.

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

Explanation:
1. Divide numerator and denominator by \( \cos^4x \): \( \frac{2\tan x\sec^2x}{\tan^4x+1} \).
2. Put \( t=\tan^2x \); then \( dt=2\tan x\sec^2x\,dx \), and \( \int\frac{dt}{1+t^2}=\tan^{-1}(\tan^2x)+c \).

Teacher's Note:
1. Dividing by cos to the 4 turns the denominator into 1 + tan to the 4.
2. The numerator becomes the derivative of tan squared.

Answer: (a) \( \frac{\sec^{3}{\left(2 x \right)}}{6} - \frac{\sec{\left(2 x \right)}}{2}+c \)

Explanation:
1. Write \( \tan^32x\sec2x=(\sec^22x-1)\sec2x\tan2x \) and put \( u=\sec2x \), \( du=2\sec2x\tan2x\,dx \).
2. \( \frac12\int(u^2-1)du=\frac{u^3}{6}-\frac u2=\frac16\sec^32x-\frac12\sec2x+c \).

Teacher's Note:
1. Keep one factor of sec tan together as the derivative of sec.
2. Convert the remaining tan squared to sec squared minus 1.

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

Explanation:
1. Put \( t=\sqrt{x^2-a^2} \), so \( x\,dx=t\,dt \) and \( x^2=t^2+a^2 \).
2. \( \int\frac{t^2}{t^2+a^2}dt=t-a\tan^{-1}\frac ta \), so the integral is \( \sqrt{x^2-a^2}-a\tan^{-1}\frac{\sqrt{x^2-a^2}}{a}+c \).

Teacher's Note:
1. Multiply numerator and denominator by the root to get x dx in the numerator.
2. Then divide t squared by t squared plus a squared.