SolveItClass 12 · NCERT

NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.6 11–20 (part 16 of 27)

  1. Integrate the functions in Exercises $\displaystyle 1$ to 22.

    Exercise 11

    xcos⁡−1x1−x2\displaystyle \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}

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    NCERT’s answer
    \(\displaystyle -\sqrt{1-x^{2}} \cos ^{-1} x+x+\mathrm{C}\)
    Substitute \(\displaystyle \theta=\cos^{-1}x\), so \(\displaystyle x=\cos\theta\), \(\displaystyle dx=-\sin\theta\,d\theta\), and \(\displaystyle \sqrt{1-x^{2}}=\sin\theta\) (positive, since \(\displaystyle \theta\in[0,\pi]\)). \[I=\int\frac{x\cos^{-1}x}{\sqrt{1-x^{2}}}\,dx = \int\frac{\cos\theta\cdot\theta}{\sin\theta}\left(-\sin\theta\right)d\theta = -\int\theta\cos\theta\,d\theta\] By parts, first function \(\displaystyle \theta\): \[\int\theta\cos\theta\,d\theta = \theta\sin\theta-\int\sin\theta\,d\theta = \theta\sin\theta+\cos\theta\] So \(\displaystyle I=-\theta\sin\theta-\cos\theta+\mathrm{C}\). Restoring \(\displaystyle \theta=\cos^{-1}x\), \(\displaystyle \sin\theta=\sqrt{1-x^{2}}\), \(\displaystyle \cos\theta=x\): \[I = -\sqrt{1-x^{2}}\,\cos^{-1}x-x+\mathrm{C}\] Final answer: \(\displaystyle -\sqrt{1-x^{2}}\,\cos^{-1}x-x+\mathrm{C}\), for \(\displaystyle -1<x<1\).

  2. Exercise 12

    xsec⁡2x\displaystyle x \sec ^{2} x

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    NCERT’s answer
    \(\displaystyle x \tan x+\log |\cos x|+\mathrm{C}\)
    Parts with first function \(\displaystyle x\) and second function \(\displaystyle \sec^{2}x\), whose integral is \(\displaystyle \tan x\). \[\int x\sec^{2}x\,dx = x\tan x-\int 1\cdot\tan x\,dx\] Using the standard result \(\displaystyle \displaystyle\int\tan x\,dx=-\log|\cos x|+\mathrm{C}\), \[= x\tan x+\log|\cos x|+\mathrm{C}\] Final answer: \(\displaystyle x\tan x+\log|\cos x|+\mathrm{C}\) (equivalently \(\displaystyle x\tan x-\log|\sec x|+\mathrm{C}\)), for \(\displaystyle x\neq(2n+1)\dfrac{\pi}{2}\).

  3. Exercise 13

    tan⁡−1x\displaystyle \tan ^{-1} x

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    NCERT’s answer
    \(\displaystyle x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^{2}\right)+C\)
    A single inverse function is integrated by parts against the second function \(\displaystyle 1\). \[\int\tan^{-1}x\,dx = \int\left(\tan^{-1}x\right)\cdot 1\,dx = \tan^{-1}x\cdot x-\int\frac{1}{1+x^{2}}\cdot x\,dx\] For the last integral put \(\displaystyle t=1+x^{2}\), \(\displaystyle dt=2x\,dx\): \[\int\frac{x}{1+x^{2}}\,dx = \frac12\int\frac{dt}{t} = \frac12\log\left(1+x^{2}\right)\] Hence \[\int\tan^{-1}x\,dx = x\tan^{-1}x-\frac12\log\left(1+x^{2}\right)+\mathrm{C}\] Final answer: \(\displaystyle x\tan^{-1}x-\dfrac12\log\left(1+x^{2}\right)+\mathrm{C}\).

  4. Exercise 14

    x(log⁡x)2\displaystyle x(\log x)^{2}

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    NCERT’s answer
    \(\displaystyle \frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+\mathrm{C}\)
    By ILATE, \(\displaystyle (\log x)^{2}\) is the first function and \(\displaystyle x\) the second; \(\displaystyle \dfrac{d}{dx}(\log x)^{2}=\dfrac{2\log x}{x}\) by the chain rule. \[\int x(\log x)^{2}dx = (\log x)^{2}\frac{x^{2}}{2}-\int\frac{2\log x}{x}\cdot\frac{x^{2}}{2}\,dx = \frac{x^{2}}{2}(\log x)^{2}-\int x\log x\,dx\] The leftover integral is Question $\displaystyle 4$: \(\displaystyle \displaystyle\int x\log x\,dx=\frac{x^{2}}{2}\log x-\frac{x^{2}}{4}\). Therefore \[\int x(\log x)^{2}dx = \frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2}\log x+\frac{x^{2}}{4}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{x^{2}}{2}(\log x)^{2}-\dfrac{x^{2}}{2}\log x+\dfrac{x^{2}}{4}+\mathrm{C}\), valid for \(\displaystyle x>0\).

  5. Exercise 15

    (x2+1)log⁡x\displaystyle \left(x^{2}+1\right) \log x

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    NCERT’s answer
    \(\displaystyle \left(\frac{x^{3}}{3}+x\right) \log x-\frac{x^{3}}{9}-x+\mathrm{C}\)
    Take \(\displaystyle \log x\) as the first function and \(\displaystyle x^{2}+1\) as the second, so the second must be integrated first: \(\displaystyle \displaystyle\int\left(x^{2}+1\right)dx=\frac{x^{3}}{3}+x\). \[\int\left(x^{2}+1\right)\log x\,dx = \log x\left(\frac{x^{3}}{3}+x\right)-\int\frac{1}{x}\left(\frac{x^{3}}{3}+x\right)dx\] \[= \left(\frac{x^{3}}{3}+x\right)\log x-\int\left(\frac{x^{2}}{3}+1\right)dx = \left(\frac{x^{3}}{3}+x\right)\log x-\frac{x^{3}}{9}-x+\mathrm{C}\] Final answer: \(\displaystyle \left(\dfrac{x^{3}}{3}+x\right)\log x-\dfrac{x^{3}}{9}-x+\mathrm{C}\), valid for \(\displaystyle x>0\).

  6. Exercise 16

    ex(sin⁡x+cos⁡x)\displaystyle e^{x}(\sin x+\cos x)

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    NCERT’s answer
    \(\displaystyle e^{x} \sin x+\mathrm{C}\)
    Use the standard result \(\displaystyle \displaystyle\int e^{x}\left[f(x)+f'(x)\right]dx = e^{x}f(x)+\mathrm{C}\). Here the bracket is \(\displaystyle \sin x+\cos x\); taking \(\displaystyle f(x)=\sin x\) gives \(\displaystyle f'(x)=\cos x\), so the integrand is exactly \(\displaystyle e^{x}\left[f(x)+f'(x)\right]\). \[\int e^{x}\left(\sin x+\cos x\right)dx = e^{x}\sin x+\mathrm{C}\] Final answer: \(\displaystyle e^{x}\sin x+\mathrm{C}\).

  7. Exercise 17

    xex(1+x)2\displaystyle \frac{x e^{x}}{(1+x)^{2}}

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    NCERT’s answer
    \(\displaystyle \frac{e^{x}}{1+x}+\mathrm{C}\)
    Aim for the form \(\displaystyle \displaystyle\int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+\mathrm{C}\), so split the rational factor by writing the numerator in terms of \(\displaystyle 1+x\): \[\frac{x}{(1+x)^{2}} = \frac{(1+x)-1}{(1+x)^{2}} = \frac{1}{1+x}-\frac{1}{(1+x)^{2}}\] With \(\displaystyle f(x)=\dfrac{1}{1+x}\) we get \(\displaystyle f'(x)=-\dfrac{1}{(1+x)^{2}}\), so the integrand is \(\displaystyle e^{x}\left[f(x)+f'(x)\right]\). \[\int\frac{x e^{x}}{(1+x)^{2}}\,dx = \frac{e^{x}}{1+x}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{e^{x}}{1+x}+\mathrm{C}\), for \(\displaystyle x\neq-1\).

  8. Exercise 18

    ex(1+sin⁡x1+cos⁡x)\displaystyle e^{x}\left(\frac{1+\sin x}{1+\cos x}\right)

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    NCERT’s answer
    \(\displaystyle e^{x} \tan \frac{x}{2}+\mathrm{C}\)
    Convert to half angles using \(\displaystyle 1+\cos x = 2\cos^{2}\dfrac{x}{2}\) and \(\displaystyle \sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\): \[\frac{1+\sin x}{1+\cos x} = \frac{1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\cos^{2}\dfrac{x}{2}} = \frac12\sec^{2}\frac{x}{2}+\tan\frac{x}{2}\] So the integrand is \(\displaystyle e^{x}\left[f(x)+f'(x)\right]\) with \(\displaystyle f(x)=\tan\dfrac{x}{2}\), since \(\displaystyle f'(x)=\dfrac12\sec^{2}\dfrac{x}{2}\). \[\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right)dx = e^{x}\tan\frac{x}{2}+\mathrm{C}\] Final answer: \(\displaystyle e^{x}\tan\dfrac{x}{2}+\mathrm{C}\), for \(\displaystyle x\neq(2n+1)\pi\).

  9. Exercise 19

    ex(1x−1x2)\displaystyle e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)

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    NCERT’s answer
    \(\displaystyle \frac{e^{x}}{x}+\mathrm{C}\)
    Match the pattern \(\displaystyle \displaystyle\int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+\mathrm{C}\). Taking \(\displaystyle f(x)=\dfrac{1}{x}\) gives \(\displaystyle f'(x)=-\dfrac{1}{x^{2}}\), and the bracket \(\displaystyle \dfrac{1}{x}-\dfrac{1}{x^{2}}\) is exactly \(\displaystyle f(x)+f'(x)\). \[\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)dx = \frac{e^{x}}{x}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{e^{x}}{x}+\mathrm{C}\), for \(\displaystyle x\neq 0\).

  10. Exercise 20

    (x−3)ex(x−1)3\displaystyle \frac{(x-3) e^{x}}{(x-1)^{3}}

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    NCERT’s answer
    \(\displaystyle \frac{e^{x}}{(x-1)^{2}}+\mathrm{C}\)
    Rewrite the numerator in terms of \(\displaystyle x-1\) so the \(\displaystyle e^{x}\left[f(x)+f'(x)\right]\) pattern appears: \(\displaystyle x-3=(x-1)-2\), hence \[\frac{x-3}{(x-1)^{3}} = \frac{(x-1)-2}{(x-1)^{3}} = \frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\] With \(\displaystyle f(x)=\dfrac{1}{(x-1)^{2}}\) we have \(\displaystyle f'(x)=-\dfrac{2}{(x-1)^{3}}\), so the integrand is \(\displaystyle e^{x}\left[f(x)+f'(x)\right]\). \[\int\frac{(x-3)e^{x}}{(x-1)^{3}}\,dx = \frac{e^{x}}{(x-1)^{2}}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{e^{x}}{(x-1)^{2}}+\mathrm{C}\), for \(\displaystyle x\neq 1\).