JEE Mathematics Indefinite Integration MCQs Set 27

Answer: (a) \( \ln|\sin x+\cos x|+C \)

Explanation:
1. Since \( \cos2x=(\cos x-\sin x)(\cos x+\sin x) \), the integrand is \( \frac{\cos x-\sin x}{\cos x+\sin x} \).
2. The numerator is the derivative of the denominator, so the integral is \( \ln|\sin x+\cos x|+C \).

Teacher's Note:
1. Factorise cos 2x as a difference of squares.
2. The numerator is the derivative of the denominator.

Answer: (b) \( \cot\alpha\,\ln\left|\frac{\sec(x+\alpha)}{\sec x}\right|+C \)

Explanation:
1. From \( \tan\alpha=\tan((x+\alpha)-x)=\frac{\tan(x+\alpha)-\tan x}{1+\tan x\tan(x+\alpha)} \) we get \( 1+\tan x\tan(x+\alpha)=\cot\alpha[\tan(x+\alpha)-\tan x] \).
2. Integrate: \( \cot\alpha\left[\ln|\sec(x+\alpha)|-\ln|\sec x|\right] \).

Teacher's Note:
1. Use the tangent subtraction formula.
2. tan integrates to ln|sec|.

Answer: (a) \( -2\ln\left(\cos\frac x2+\sqrt{\cos^2\frac x2-\frac12}\right)+C \)

Explanation:
1. Write \( \sqrt{\sec x-1}=\frac{\sqrt{1-\cos x}}{\sqrt{\cos x}}=\frac{\sqrt2\sin(x/2)}{\sqrt{2\cos^2(x/2)-1}} \).
2. Put \( t=\cos\frac x2 \), \( dt=-\frac12\sin\frac x2dx \): \( -2\int\frac{dt}{\sqrt{t^2-1/2}} \).
3. This equals \( -2\ln\left(t+\sqrt{t^2-\tfrac12}\right) \).

Teacher's Note:
1. Use half-angle forms for 1 - cos x and cos x.
2. The final integral is of the form 1/sqrt(t^2 - a^2).

Answer: (b) \( \sqrt2\left(\sqrt{\tan x}+\frac15\tan^{5/2}x\right)+C \)

Explanation:
1. Since \( \sin2x=2\sin x\cos x \), the integrand is \( \frac{1}{\sqrt2\cos^{7/2}x\sqrt{\sin x}}=\frac{\sec^2x\,(1+\tan^2x)}{\sqrt2\sqrt{\tan x}} \).
2. With \( t=\tan x \): \( \frac1{\sqrt2}\int(t^{-1/2}+t^{3/2})dt \) which is \( \frac{1}{\sqrt2}\left(2\sqrt t+\frac25t^{5/2}\right)=\sqrt2\left(\sqrt t+\frac15t^{5/2}\right) \).

Teacher's Note:
1. Divide by cos^4 x to bring in tan x and sec^2 x.
2. sec^2 x = 1 + tan^2 x.

Answer: (d) \( x-\frac{\sqrt2}{2}\tan^{-1}(\sqrt2\tan x)+C \)

Explanation:
1. Write the integrand as \( 1-\frac1{1+\sin^2x} \).
2. For the second part, divide by \( \cos^2x \): \( \int\frac{\sec^2x\,dx}{\sec^2x+\tan^2x}=\int\frac{dt}{1+2t^2}=\frac1{\sqrt2}\tan^{-1}(\sqrt2t) \) with \( t=\tan x \).
3. So the result is \( x-\frac{\sqrt2}2\tan^{-1}(\sqrt2\tan x) \).

Teacher's Note:
1. Divide numerator and denominator by cos^2 x.
2. sec^2 x = 1 + tan^2 x.

Answer: (b) \( \frac12\ln|\sin x-\cos x|+\frac12x+C \)

Explanation:
1. The integrand is \( \frac{\sin x}{\sin x-\cos x} \).
2. Write \( \sin x=\tfrac12[(\sin x-\cos x)+(\sin x+\cos x)] \).
3. Then \( \frac12\int dx+\frac12\int\frac{\sin x+\cos x}{\sin x-\cos x}dx=\frac x2+\frac12\ln|\sin x-\cos x| \).

Teacher's Note:
1. Express the numerator as a combination of the denominator and its derivative.
2. The derivative of sin x - cos x is sin x + cos x.

Answer: (c) \( \frac1{2\sqrt3}\ln\left|\frac{\sqrt3+\sin x-\cos x}{\sqrt3-(\sin x-\cos x)}\right|+\tan^{-1}(\sin x+\cos x)+C \)

Explanation:
1. The integrand is \( \frac{\cos x}{\sin x\cos x+1} \). Split \( \cos x=\tfrac12[(\cos x+\sin x)+(\cos x-\sin x)] \).
2. For the part with \( (\cos x-\sin x)dx=du \), where \( u=\sin x+\cos x \) and \( \sin x\cos x+1=\frac{u^2+1}{2} \): \( \int\frac{du}{u^2+1}=\tan^{-1}u \).
3. For the part with \( (\cos x+\sin x)dx=dv \), where \( v=\sin x-\cos x \) and \( \sin x\cos x+1=\frac{3-v^2}{2} \): \( \int\frac{dv}{3-v^2}=\frac1{2\sqrt3}\ln\left|\frac{\sqrt3+v}{\sqrt3-v}\right| \).

Teacher's Note:
1. Split cos x into (cos x + sin x) and (cos x - sin x) parts.
2. Use the two substitutions u = sin x + cos x and v = sin x - cos x.

Answer: (c) \( \frac1{\cos(a-b)}\ln\left|\frac{\sin(x-a)}{\cos(x-b)}\right|+C \)

Explanation:
1. Note \( \cos(a-b)=\cos((x-b)-(x-a))=\cos(x-b)\cos(x-a)+\sin(x-b)\sin(x-a) \).
2. Divide by \( \sin(x-a)\cos(x-b) \): \( \frac{\cos(a-b)}{\sin(x-a)\cos(x-b)}=\cot(x-a)+\tan(x-b) \).
3. Integrate: \( \ln|\sin(x-a)|-\ln|\cos(x-b)| \), then divide by \( \cos(a-b) \).

Teacher's Note:
1. Express cos(a - b) through the two angles x - a and x - b.
2. Then the integrand splits into cot and tan.

Answer: (c) \( -\ln|\sec x|-\frac12\ln|\sec2x|+\frac13\ln|\sec3x|+C \)

Explanation:
1. Since \( \tan3x=\frac{\tan2x+\tan x}{1-\tan x\tan2x} \), we get \( \tan x\tan2x\tan3x=\tan3x-\tan2x-\tan x \).
2. Integrate each: \( \frac13\ln|\sec3x|-\frac12\ln|\sec2x|-\ln|\sec x| \).

Teacher's Note:
1. Use tan 3x = tan(2x + x).
2. The integral of tan kx is (1/k) ln|sec kx|.

Answer: (d) \( -\frac{\cos32x}{1024}+C \)

Explanation:
1. Repeatedly use \( \sin\theta\cos\theta=\frac12\sin2\theta \) to get \( \frac{\sin32x}{32} \).
2. Integrate: \( -\frac{\cos32x}{32\cdot32}=-\frac{\cos32x}{1024} \).

Teacher's Note:
1. Apply the double-angle identity five times.
2. 1024 = 32 x 32.

Answer: (d) \( \frac12a^2\sin^{-1}\frac{x^2}{a^2}+\frac12\sqrt{a^4-x^4}+C \)

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

Teacher's Note:
1. Multiply numerator and denominator by a^2 - t.
2. Split into an arcsine and an algebraic part.

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

Explanation:
1. Rationalise: \( \sqrt{\frac{x-1}{x+1}}=\frac{x-1}{\sqrt{x^2-1}} \).
2. The integrand becomes \( \frac{1}{x\sqrt{x^2-1}}-\frac{1}{x^2\sqrt{x^2-1}} \).
3. The first part is \( \sec^{-1}x \); with \( x=\sec\theta \) the second is \( \int\cos\theta\,d\theta=\frac{\sqrt{x^2-1}}{x} \).

Teacher's Note:
1. Multiply by (x - 1)/(x - 1) under the root.
2. Use x = sec theta for the second integral.

Answer: (a) \( \frac23\sqrt{1+\ln|x|}\,(\ln|x|-2)+C \)

Explanation:
1. Put \( u=\ln|x| \): \( \int\frac{u}{\sqrt{1+u}}du \).
2. With \( s=1+u \): \( \int(s^{1/2}-s^{-1/2})ds=\frac23s^{3/2}-2s^{1/2}=\frac23\sqrt s\,(s-3) \).
3. Since \( s-3=u-2 \), the result is \( \frac23\sqrt{1+\ln|x|}(\ln|x|-2) \).

Teacher's Note:
1. Two substitutions: u = ln|x| then s = 1 + u.
2. Factor out sqrt(s).

Answer: (b) \( \frac43\left(\frac{x-1}{x+2}\right)^{1/4}+C \)

Explanation:
1. Write the integrand as \( \left(\frac{x-1}{x+2}\right)^{-3/4}\frac{1}{(x+2)^2} \).
2. Put \( t=\frac{x-1}{x+2} \), \( dt=\frac{3\,dx}{(x+2)^2} \).
3. Then \( \frac13\int t^{-3/4}dt=\frac43t^{1/4} \).

Teacher's Note:
1. Factor (x + 2)^2 out so the ratio appears.
2. The derivative of (x - 1)/(x + 2) is 3/(x + 2)^2.

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

Explanation:
1. Put \( x=\cos^2\theta \), so \( \sqrt x=\cos\theta \), \( \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\tan\frac\theta2 \) and \( dx=-2\sin\theta\cos\theta\,d\theta \).
2. The integral is \( -2\int\tan\frac\theta2\sin\theta\cos\theta\,d\theta=-2\int(1-\cos\theta)\cos\theta\,d\theta \) because \( \tan\frac\theta2\sin\theta=1-\cos\theta \).
3. This equals \( -2\sin\theta+\theta+\sin\theta\cos\theta \), i.e. \( \sqrt x\sqrt{1-x}-2\sqrt{1-x}+\cos^{-1}\sqrt x \).

Teacher's Note:
1. Substitute x = cos^2 theta.
2. Verify the answer by differentiation.

Answer: (a) \( (a+x)\tan^{-1}\sqrt{\frac xa}-\sqrt{ax}+C \)

Explanation:
1. Note \( \sin^{-1}\sqrt{\frac x{a+x}}=\tan^{-1}\sqrt{\frac xa} \).
2. By parts: \( x\tan^{-1}\sqrt{x/a}-\int x\cdot\frac{\sqrt a}{2\sqrt x(a+x)}dx \).
3. The remaining integral gives \( \sqrt{ax}-a\tan^{-1}\sqrt{x/a} \), so the result is \( (a+x)\tan^{-1}\sqrt{x/a}-\sqrt{ax} \).

Teacher's Note:
1. Convert the inverse sine to an inverse tangent.
2. Then integrate by parts.

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

Explanation:
1. Put \( t=\sqrt x \): the integral is \( 2\int e^t(t^2+t)dt \).
2. Using \( \int e^t(t^2+t)dt=e^t(t^2-t+1) \) we obtain \( 2e^{\sqrt x}(x-\sqrt x+1) \).

Teacher's Note:
1. The factor 1/sqrt x is absorbed by dx = 2 sqrt x dt.
2. Integrate t^2 e^t by parts twice.

Answer: (d) 1

Explanation:
1. Put \( u=\ln x \): \( x^{\ln x}=e^{u^2} \) and \( \frac{dx}{x}=du \).
2. Then \( I=2\int u^3e^{u^2}du=\int we^w\,dw=(w-1)e^w \) with \( w=u^2 \).
3. So \( I=x^{\ln x}(\ln x)^2-x^{\ln x} \), giving A = B = 1.

Teacher's Note:
1. Write x^(ln x) as e^((ln x)^2).
2. Put w = u^2 and integrate w e^w by parts.

Answer: (a) \( e^{\tan\theta}\cos\theta+C \)

Explanation:
1. Differentiate \( e^{\tan\theta}\cos\theta \): \( e^{\tan\theta}(\sec^2\theta\cos\theta-\sin\theta)=e^{\tan\theta}(\sec\theta-\sin\theta) \).
2. So the integral is \( e^{\tan\theta}\cos\theta+C \).

Teacher's Note:
1. Use the form e^f (f' g + g') pattern with g = cos theta.
2. Check by differentiating the answer.

Answer: (a) \( x\ln(1+\sin x)+C \)

Explanation:
1. Note \( \tan\left(\frac\pi4-\frac x2\right)=\frac{\cos x}{1+\sin x} \).
2. Then \( \frac{d}{dx}[x\ln(1+\sin x)]=\ln(1+\sin x)+\frac{x\cos x}{1+\sin x} \), which equals the integrand.
3. So the integral is \( x\ln(1+\sin x)+C \).

Teacher's Note:
1. Simplify the tangent of the difference.
2. The integrand has the form f + x f'.

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

Explanation:
1. By parts with \( u=\ln(x+\sqrt{1+x^2}) \), \( dv=\frac{x\,dx}{\sqrt{1+x^2}} \), \( v=\sqrt{1+x^2} \).
2. Since \( du=\frac{dx}{\sqrt{1+x^2}} \), we get \( \sqrt{1+x^2}\ln(\cdot)-\int dx \).
3. Hence \( \sqrt{1+x^2}\ln\left(x+\sqrt{1+x^2}\right)-x+C \).

Teacher's Note:
1. Choose the logarithm as the first function.
2. d/dx ln(x + sqrt(1 + x^2)) = 1/sqrt(1 + x^2).

Answer: (a) f(x) = tan^-1 x and A = -1

Explanation:
1. By parts with \( u=\tan^{-1}x \) and \( dv=\frac{x\,dx}{\sqrt{1+x^2}} \): \( \sqrt{1+x^2}\tan^{-1}x-\int\frac{\sqrt{1+x^2}}{1+x^2}dx \).
2. The last integral is \( \int\frac{dx}{\sqrt{1+x^2}}=\ln|x+\sqrt{x^2+1}| \).
3. So \( f(x)=\tan^{-1}x \) and \( A=-1 \).

Teacher's Note:
1. Integrate x/sqrt(1 + x^2) first.
2. Then the remaining integral is standard.

Answer: (b) \( (x+1)+2\sqrt{x+1}-2\ln|x+2|-2\tan^{-1}\sqrt{x+1}+C \)

Explanation:
1. Put \( t=\sqrt{x+1} \), \( dx=2t\,dt \), \( x+2=t^2+1 \), \( x=t^2-1 \).
2. The integral is \( 2\int\frac{t^3+t^2-t}{t^2+1}dt \). Dividing: \( \frac{t^3+t^2-t}{t^2+1}=t+1-\frac{2t+1}{t^2+1} \).
3. Integrating: \( t^2+2t-2\ln(t^2+1)-2\tan^{-1}t \), which in x is the stated result.

Teacher's Note:
1. Substitute t = sqrt(x + 1).
2. Verify by differentiating.

Answer: (d) \( -2\sin^{-1}\left(\frac{\cos(x/2)}{\cos(\alpha/2)}\right)+C \)

Explanation:
1. Use \( 1-\cos x=2\sin^2\frac x2 \) and \( \cos\alpha-\cos x=2\cos^2\frac\alpha2-2\cos^2\frac x2 \).
2. Put \( t=\cos\frac x2 \), \( dt=-\frac12\sin\frac x2dx \): the integrand becomes \( \frac{\sin(x/2)\,dx}{\sqrt{\cos^2(\alpha/2)-\cos^2(x/2)}}=\frac{-2\,dt}{\sqrt{c^2-t^2}} \) with \( c=\cos\frac\alpha2 \).
3. So the result is \( -2\sin^{-1}\frac tc \).

Teacher's Note:
1. Convert both numerator and denominator to half angles.
2. The final integral is of the form 1/sqrt(c^2 - t^2).