JEE Mathematics Indefinite Integration MCQs Set 29

Answer: (a) \(x^4+\frac1{x^3}-\frac{129}{8}\)

Explanation:
1. Integrating, \(f(x)=x^4+\frac{1}{x^3}+C\).
2. \(f(2)=16+\frac18+C=0\), so \(C=-\frac{129}{8}\).
3. Hence \(f(x)=x^4+\frac1{x^3}-\frac{129}{8}\).

Teacher's Note:
1. Use the given value to fix the constant of integration. Write the antiderivative of \(x^{-4}\) carefully.

Answer: (a) \(\frac25(x+2)^{5/2}-\frac43(x+2)^{3/2}+C\)

Explanation:
1. Put \(u=x+2\), so \(x=u-2\) and \(dx=du\).
2. \(\int(u-2)u^{1/2}du=\frac25u^{5/2}-\frac43u^{3/2}+C\).

Teacher's Note:
1. Write the other factor in terms of the new variable. Then integrate term by term.

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

Explanation:
1. \(e^{n\log x}=x^n\), so the integrand is \(\frac{x^6-x^5}{x^4-x^3}=\frac{x^5(x-1)}{x^3(x-1)}=x^2\).
2. \(\int x^2dx=\frac{x^3}{3}+C\).

Teacher's Note:
1. Simplify exponentials of logarithms first. Cancel the common factor before integrating.

Answer: (b) \(\tan\frac x2+C\)

Explanation:
1. \(\csc^2x=\frac{1}{\sin^2x}=\frac{1}{(1-\cos x)(1+\cos x)}\), so the integrand is \(\frac{1}{1+\cos x}=\frac{1}{2\cos^2(x/2)}\).
2. \(\int\frac12\sec^2\frac x2dx=\tan\frac x2+C\).

Teacher's Note:
1. Use \(1+\cos x=2\cos^2rac x2\). The integral of \(\sec^2\) is the tangent.

Answer: (c) \(2\log|\sqrt x-1|+C\)

Explanation:
1. Put \(\sqrt x=t\), \(dx=2t\,dt\).
2. \(\int\frac{2t\,dt}{t^2-t}=2\int\frac{dt}{t-1}=2\log|t-1|+C\).
3. So the answer is \(2\log|\sqrt x-1|+C\).

Teacher's Note:
1. Substitute for the root; the factor t cancels neatly.

Answer: (c) \(\frac{(\log x)^{1-m}}{1-m}+C\)

Explanation:
1. Put \(t=\log x\), \(dt=\frac{dx}{x}\).
2. \(\int t^{-m}dt=\frac{t^{1-m}}{1-m}+C\) (for \(m\ne1\)).

Teacher's Note:
1. Recognise \(rac{dx}{x}\) as \(d(\log x)\).

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

Explanation:
1. Put \(t=\tan^{-1}x\), so \(dt=\frac{dx}{1+x^2}\).
2. \(\int e^tdt=e^t+C=e^{\tan^{-1}x}+C\).

Teacher's Note:
1. The factor \(rac{1}{1+x^2}\) signals the substitution \(t= an^{-1}x\).

Answer: (a) \(\log|e^x+e^{-x}|+C\)

Explanation:
1. Divide numerator and denominator by \(e^x\): \(\frac{e^x-e^{-x}}{e^x+e^{-x}}\).
2. The numerator is the derivative of the denominator, so the integral is \(\log|e^x+e^{-x}|+C\).

Teacher's Note:
1. This is \( anh x\). Spot the form \(rac{f'}{f}\).

Answer: (b) \(\frac12\log|e^{2x}+e^{-2x}|+C\)

Explanation:
1. The derivative of the denominator is \(2e^{2x}-2e^{-2x}=2(e^{2x}-e^{-2x})\).
2. So \(\int\frac{f'}{2f}dx=\frac12\log|e^{2x}+e^{-2x}|+C\).

Teacher's Note:
1. Adjust the constant factor when the numerator is only a multiple of the derivative.

Answer: (c) \(\frac12(\sin^{-1}x)^2+C\)

Explanation:
1. Put \(t=\sin^{-1}x\), \(dt=\frac{dx}{\sqrt{1-x^2}}\).
2. \(\int t\,dt=\frac{t^2}{2}+C=\frac12(\sin^{-1}x)^2+C\).

Teacher's Note:
1. Whenever \(rac1{\sqrt{1-x^2}}\) appears with \(\sin^{-1}x\), substitute \(t=\sin^{-1}x\).

Answer: (a) \(\frac12\log|3\cos x+2\sin x|+C\)

Explanation:
1. The derivative of the denominator is \(-6\sin x+4\cos x=2(2\cos x-3\sin x)\).
2. So the integral is \(\frac12\log|6\cos x+4\sin x|+C=\frac12\log|3\cos x+2\sin x|+C'\).

Teacher's Note:
1. The numerator is half the derivative of the denominator. The constant \(\log2\) merges with C.

Answer: (b) \(2\sin\sqrt x+C\)

Explanation:
1. Put \(t=\sqrt x\), so \(\frac{dx}{\sqrt x}=2dt\).
2. \(\int2\cos t\,dt=2\sin t+C=2\sin\sqrt x+C\).

Teacher's Note:
1. The factor \(rac1{\sqrt x}dx\) equals \(2\,d(\sqrt x)\).

Answer: (d) \(\frac12(\log\sin x)^2+C\)

Explanation:
1. Put \(t=\log\sin x\), so \(dt=\cot x\,dx\).
2. \(\int t\,dt=\frac{t^2}{2}+C=\frac12(\log\sin x)^2+C\).

Teacher's Note:
1. The derivative of \(\log\sin x\) is \(\cot x\), which is already present.

Answer: (a) \(\frac12[x-\log|\cos x-\sin x|]+C\)

Explanation:
1. \(\frac{1}{1-\tan x}=\frac{\cos x}{\cos x-\sin x}=\frac12\left[1+\frac{\cos x+\sin x}{\cos x-\sin x}\right]\).
2. \(\int\frac{\cos x+\sin x}{\cos x-\sin x}dx=-\log|\cos x-\sin x|\).
3. So the result is \(\frac12[x-\log|\cos x-\sin x|]+C\).

Teacher's Note:
1. Split the numerator as a combination of the denominator and its derivative.

Answer: (d) \(\log|10^x+x^{10}|+C\)

Explanation:
1. The derivative of \(x^{10}+10^x\) is \(10x^9+10^x\log_e10\), which is the numerator.
2. So the integral is \(\log|10^x+x^{10}|+C\).

Teacher's Note:
1. The form \(rac{f'(x)}{f(x)}\) integrates to \(\log|f(x)|\).

Answer: (a) \(-\frac12\cos(2x+1)+\frac{\cos^3(2x+1)}{6}+C\)

Explanation:
1. Let \(\theta=2x+1\), \(dx=\frac{d\theta}{2}\). \(\sin^3\theta=\sin\theta(1-\cos^2\theta)\).
2. \(\frac12\int\sin\theta(1-\cos^2\theta)d\theta=\frac12\left[-\cos\theta+\frac{\cos^3\theta}{3}\right]\).
3. So the answer is \(-\frac12\cos(2x+1)+\frac{\cos^3(2x+1)}{6}+C\).

Teacher's Note:
1. Odd powers of sine are integrated by saving one sine and converting the rest to cosine.

Answer: (c) \(-\frac{1}{\sin x+\cos x}+C\)

Explanation:
1. \(1+\sin2x=(\sin x+\cos x)^2\). Put \(t=\sin x+\cos x\), \(dt=(\cos x-\sin x)dx\).
2. \(\int\frac{dt}{t^2}=-\frac1t+C=-\frac{1}{\sin x+\cos x}+C\).

Teacher's Note:
1. Recognise \(1+\sin2x\) as a perfect square. The numerator is then the derivative of the base.

Answer: (a) \(\frac16\sec^32x-\frac12\sec2x+C\)

Explanation:
1. \(\tan^32x\sec2x=(\sec^22x-1)\sec2x\tan2x\). Put \(u=\sec2x\), \(du=2\sec2x\tan2x\,dx\).
2. \(\frac12\int(u^2-1)du=\frac12\left(\frac{u^3}{3}-u\right)\).
3. So the answer is \(\frac16\sec^32x-\frac12\sec2x+C\).

Teacher's Note:
1. Save one factor of \(\sec an\) for the differential and write the rest in terms of sec.

Answer: (b) \(\tan(xe^x)+C\)

Explanation:
1. Put \(t=xe^x\), so \(dt=e^x(1+x)dx\).
2. \(\int\sec^2t\,dt=\tan t+C=\tan(xe^x)+C\).

Teacher's Note:
1. Look for a function whose derivative is the other factor of the integrand.

Answer: (d) \(\frac{3}{2\sqrt2}\tan^{-1}(\sqrt2x^2)+C\)

Explanation:
1. Put \(u=x^2\), \(du=2x\,dx\). The integral becomes \(\frac32\int\frac{du}{1+2u^2}\).
2. \(\int\frac{du}{1+2u^2}=\frac{1}{\sqrt2}\tan^{-1}(\sqrt2u)\).
3. The result is \(\frac{3}{2\sqrt2}\tan^{-1}(\sqrt2x^2)+C\).

Teacher's Note:
1. Use \(\intrac{du}{1+k^2u^2}=rac1k an^{-1}(ku)\).

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

Explanation:
1. \(\frac{1}{e^x-1}=\frac{e^{-x}}{1-e^{-x}}\). Put \(t=1-e^{-x}\), \(dt=e^{-x}dx\).
2. \(\int\frac{dt}{t}=\log|1-e^{-x}|=\log\left|\frac{e^x-1}{e^x}\right|+C\).

Teacher's Note:
1. Multiply by \(rac{e^{-x}}{e^{-x}}\) to expose the derivative of the denominator.

Answer: (c) \(-(1+x^{-4})^{1/4}+C\)

Explanation:
1. Write \((x^4+1)^{3/4}=x^3(1+x^{-4})^{3/4}\), so the integrand is \(\frac{1}{x^5(1+x^{-4})^{3/4}}\).
2. Put \(t=1+x^{-4}\), \(dt=-4x^{-5}dx\). The integral is \(-\frac14\int t^{-3/4}dt=-t^{1/4}\).
3. So the answer is \(-(1+x^{-4})^{1/4}+C\).

Teacher's Note:
1. Factor the highest power of x out of the bracket to create a substitution.

Answer: (a) \(\tan^{-1}e^x+C\)

Explanation:
1. \(\frac{1}{e^x+e^{-x}}=\frac{e^x}{e^{2x}+1}\). Put \(t=e^x\).
2. \(\int\frac{dt}{1+t^2}=\tan^{-1}t=\tan^{-1}e^x+C\).

Teacher's Note:
1. Multiply numerator and denominator by \(e^x\) to get a standard form.

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

Explanation:
1. \(e^x+e^{-x}+2=\frac{(e^x+1)^2}{e^x}\), so the integrand is \(\frac{e^x}{(e^x+1)^2}\).
2. Put \(t=e^x+1\): \(\int\frac{dt}{t^2}=-\frac1t=-\frac{1}{e^x+1}+C\).

Teacher's Note:
1. The denominator is a perfect square: \((e^{x/2}+e^{-x/2})^2\).

Answer: (d) \(\frac23\sin^{-1}(x^{3/2})+C\)

Explanation:
1. Put \(t=x^{3/2}\), so \(dt=\frac32x^{1/2}dx\) and \(x^3=t^2\).
2. \(\int\frac{x^{1/2}dx}{\sqrt{1-x^3}}=\frac23\int\frac{dt}{\sqrt{1-t^2}}=\frac23\sin^{-1}t\).
3. The answer is \(\frac23\sin^{-1}(x^{3/2})+C\).

Teacher's Note:
1. Choose the substitution so that \(x^3\) becomes a perfect square.

Answer: (a) \(\frac{1}{\log3}\log|3^x+\sqrt{9^x-1}|+C\)

Explanation:
1. Put \(t=3^x\), \(dt=3^x\log3\,dx\). Since \(9^x=t^2\), the integral is \(\frac{1}{\log3}\int\frac{dt}{\sqrt{t^2-1}}\).
2. \(\int\frac{dt}{\sqrt{t^2-1}}=\log|t+\sqrt{t^2-1}|\).
3. The result is \(\frac{1}{\log3}\log|3^x+\sqrt{9^x-1}|+C\).

Teacher's Note:
1. The factor \(\log 3\) appears when differentiating \(3^x\); do not leave it out.

Answer: (d) \(\log\left[\tan^{-1}\left(x+\frac1x\right)\right]+C\)

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

Teacher's Note:
1. Divide by \(x^2\) to make the expression depend only on \(x+rac1x\).

Answer: (d) \(\sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C\)

Explanation:
1. \(8+3x-x^2=\frac{41}{4}-\left(x-\frac32\right)^2\).
2. \(\int\frac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}\frac ua\) with \(a=\frac{\sqrt{41}}{2}\), \(u=x-\frac32\).
3. The result is \(\sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C\).

Teacher's Note:
1. Complete the square under the root and use the standard sine-inverse form.

Answer: (b) \(\frac12\sin^{-1}\left(\frac{8x-9}{9}\right)+C\)

Explanation:
1. \(9x-4x^2=\frac{81}{16}-\left(2x-\frac94\right)^2\).
2. With \(u=2x-\frac94\), \(du=2dx\) and \(a=\frac94\): \(\frac12\sin^{-1}\frac{u}{a}=\frac12\sin^{-1}\frac{8x-9}{9}\).

Teacher's Note:
1. The coefficient of \(x^2\) is 4, so the substitution brings a factor \(rac12\).

Answer: (a) \(\frac{x+2}{2}\sqrt{x^2+4x+6}+\log\left|(x+2)+\sqrt{x^2+4x+6}\right|+C\)

Explanation:
1. \(x^2+4x+6=(x+2)^2+2\).
2. \(\int\sqrt{u^2+a^2}du=\frac u2\sqrt{u^2+a^2}+\frac{a^2}{2}\log|u+\sqrt{u^2+a^2}|\) with \(a^2=2\).
3. The result is \(\frac{x+2}{2}\sqrt{x^2+4x+6}+\log|(x+2)+\sqrt{x^2+4x+6}|+C\).

Teacher's Note:
1. Complete the square, then apply the standard formula for \(\int\sqrt{u^2+a^2}\).

Answer: (c) \(\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\frac{2x-3}{\sqrt{13}}+C\)

Explanation:
1. \(1+3x-x^2=\frac{13}{4}-\left(x-\frac32\right)^2\).
2. \(\int\sqrt{a^2-u^2}du=\frac u2\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}\frac ua\) with \(a^2=\frac{13}{4}\), \(u=x-\frac32\).
3. This gives \(\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\frac{2x-3}{\sqrt{13}}+C\).

Teacher's Note:
1. Remember the factor \(rac{a^2}{2}\) in front of the inverse sine.