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NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.2 21–30 (part 5 of 27)

  1. Integrate the functions in Exercises $\displaystyle 1$ to $\displaystyle 37$:

    Exercise 21

    tan⁡2(2x−3)\displaystyle \tan ^{2}(2 x-3)

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \tan (2 x-3)-x+C\)
    There is no direct formula for \(\displaystyle \int\tan^{2}\theta\,d\theta\); convert it first using the identity \(\displaystyle \sec^{2}\theta=1+\tan^{2}\theta\), i.e. \(\displaystyle \tan^{2}\theta=\sec^{2}\theta-1\), with \(\displaystyle \theta=2x-3\): \[\int\tan^{2}(2x-3)\,dx=\int\left[\sec^{2}(2x-3)-1\right]dx.\] With \(\displaystyle t=2x-3\), \(\displaystyle dt=2\,dx\), and \(\displaystyle \int\sec^{2}t\,dt=\tan t\), \[=\frac{1}{2}\tan(2x-3)-x+\mathrm{C}.\] Hence \[\int\tan^{2}(2x-3)\,dx=\frac{1}{2}\tan(2x-3)-x+\mathrm{C}.\]

  2. Exercise 22

    sec⁡2(7−4x)\displaystyle \sec ^{2}(7-4 x)

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    NCERT’s answer
    \(\displaystyle -\frac{1}{4} \tan (7-4 x)+\mathrm{C}\)
    Put \[t=7-4x\quad\Rightarrow\quad dt=-4\,dx,\ \text{ i.e. }\ dx=-\frac{dt}{4}.\] The negative sign from the coefficient of \(\displaystyle x\) is the usual slip here. Using \(\displaystyle \int\sec^{2}t\,dt=\tan t+\mathrm{C}\), \[\int\sec^{2}(7-4x)\,dx=-\frac{1}{4}\int\sec^{2}t\,dt=-\frac{1}{4}\tan t+\mathrm{C}.\] Hence \[\int\sec^{2}(7-4x)\,dx=-\frac{1}{4}\tan(7-4x)+\mathrm{C}.\]

  3. Exercise 23

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

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\left(\sin ^{-1} x\right)^{2}+\mathrm{C}\)
    Recall \(\displaystyle \dfrac{d}{dx}\left(\sin^{-1}x\right)=\dfrac{1}{\sqrt{1-x^{2}}}\), so put \[t=\sin^{-1}x\quad\Rightarrow\quad dt=\frac{dx}{\sqrt{1-x^{2}}}\qquad(-1<x<1).\] Then \[\int\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}\,dx=\int t\,dt=\frac{t^{2}}{2}+\mathrm{C}.\] Hence \[\int\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}\,dx=\frac{\left(\sin^{-1}x\right)^{2}}{2}+\mathrm{C},\qquad -1<x<1.\]

  4. Exercise 24

    2cos⁡x−3sin⁡x6cos⁡x+4sin⁡x\displaystyle \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \log |2 \sin x+3 \cos x|+C\)
    Differentiate the denominator and compare with the numerator: \[\frac{d}{dx}\left(6\cos x+4\sin x\right)=-6\sin x+4\cos x=2\left(2\cos x-3\sin x\right).\] So the numerator is exactly half the derivative of the denominator. Put \[t=6\cos x+4\sin x\quad\Rightarrow\quad dt=2(2\cos x-3\sin x)\,dx.\] Then \[\int\frac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx=\frac{1}{2}\int\frac{dt}{t}=\frac{1}{2}\log|t|+\mathrm{C}.\] Hence \[\int\frac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx=\frac{1}{2}\log\left|6\cos x+4\sin x\right|+\mathrm{C},\] which (taking \(\displaystyle 2\) out of the logarithm into the constant) may also be written as \(\displaystyle \dfrac{1}{2}\log\left|3\cos x+2\sin x\right|+\mathrm{C}\).

  5. Exercise 25

    1cos⁡2x(1−tan⁡x)2\displaystyle \frac{1}{\cos ^{2} x(1-\tan x)^{2}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{(1-\tan x)}+\mathrm{C}\)
    Write \(\displaystyle \dfrac{1}{\cos^{2}x}=\sec^{2}x\), so that \[\int\frac{dx}{\cos^{2}x(1-\tan x)^{2}}=\int\frac{\sec^{2}x}{(1-\tan x)^{2}}\,dx.\] Put \[t=1-\tan x\quad\Rightarrow\quad dt=-\sec^{2}x\,dx,\ \text{ i.e. }\ \sec^{2}x\,dx=-dt.\] Then \[\int\frac{\sec^{2}x}{(1-\tan x)^{2}}\,dx=-\int\frac{dt}{t^{2}}=-\left(-\frac{1}{t}\right)+\mathrm{C}=\frac{1}{t}+\mathrm{C}.\] Hence \[\int\frac{1}{\cos^{2}x(1-\tan x)^{2}}\,dx=\frac{1}{1-\tan x}+\mathrm{C}.\]

  6. Exercise 26

    cos⁡xx\displaystyle \frac{\cos \sqrt{x}}{\sqrt{x}}

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    NCERT’s answer
    \(\displaystyle 2 \sin \sqrt{x}+\mathrm{C}\)
    Put \[t=\sqrt{x}\quad\Rightarrow\quad dt=\frac{1}{2\sqrt{x}}\,dx,\ \text{ i.e. }\ \frac{dx}{\sqrt{x}}=2\,dt\qquad(x>0).\] Then \[\int\frac{\cos\sqrt{x}}{\sqrt{x}}\,dx=2\int\cos t\,dt=2\sin t+\mathrm{C}.\] Hence \[\int\frac{\cos\sqrt{x}}{\sqrt{x}}\,dx=2\sin\sqrt{x}+\mathrm{C},\qquad x>0.\]

  7. Exercise 27

    sin⁡2xcos⁡2x\displaystyle \sqrt{\sin 2 x} \cos 2 x

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    NCERT’s answer
    \(\displaystyle \frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+\mathrm{C}\)
    Here \(\displaystyle \cos 2x\,dx\) is proportional to the differential of \(\displaystyle \sin 2x\). Put \[t=\sin 2x\quad\Rightarrow\quad dt=2\cos 2x\,dx,\ \text{ i.e. }\ \cos 2x\,dx=\frac{dt}{2}.\] Then \[\int\sqrt{\sin 2x}\,\cos 2x\,dx=\frac{1}{2}\int t^{1/2}\,dt=\frac{1}{2}\cdot\frac{2}{3}t^{3/2}+\mathrm{C}=\frac{1}{3}t^{3/2}+\mathrm{C}.\] Hence \[\int\sqrt{\sin 2x}\,\cos 2x\,dx=\frac{1}{3}\left(\sin 2x\right)^{3/2}+\mathrm{C},\] valid on any interval where \(\displaystyle \sin 2x\geq0\) (so that the square root is defined).

  8. Exercise 28

    cos⁡x1+sin⁡x\displaystyle \frac{\cos x}{\sqrt{1+\sin x}}

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    NCERT’s answer
    \(\displaystyle 2 \sqrt{1+\sin x}+\mathrm{C}\)
    Put \[t=1+\sin x\quad\Rightarrow\quad dt=\cos x\,dx.\] Then \[\int\frac{\cos x}{\sqrt{1+\sin x}}\,dx=\int\frac{dt}{\sqrt{t}}=\int t^{-1/2}\,dt=2t^{1/2}+\mathrm{C}.\] Hence \[\int\frac{\cos x}{\sqrt{1+\sin x}}\,dx=2\sqrt{1+\sin x}+\mathrm{C},\] taken on an interval where \(\displaystyle 1+\sin x>0\).

  9. Exercise 29

    cot⁡xlog⁡sin⁡x\displaystyle \cot x \log \sin x

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}(\log \sin x)^{2}+\mathrm{C}\)
    Note that \(\displaystyle \dfrac{d}{dx}\left(\log\sin x\right)=\dfrac{\cos x}{\sin x}=\cot x\), so the whole factor \(\displaystyle \cot x\,dx\) is the differential of the logarithm. Put \[t=\log\sin x\quad\Rightarrow\quad dt=\cot x\,dx\qquad(\sin x>0).\] Then \[\int\cot x\,\log\sin x\,dx=\int t\,dt=\frac{t^{2}}{2}+\mathrm{C}.\] Hence \[\int\cot x\,\log\sin x\,dx=\frac{\left(\log\sin x\right)^{2}}{2}+\mathrm{C},\qquad \sin x>0.\]

  10. Exercise 30

    sin⁡x1+cos⁡x\displaystyle \frac{\sin x}{1+\cos x}

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    NCERT’s answer
    \(\displaystyle -\log |1+\cos x|+\mathrm{C}\)
    Put \[t=1+\cos x\quad\Rightarrow\quad dt=-\sin x\,dx,\ \text{ i.e. }\ \sin x\,dx=-dt.\] Then \[\int\frac{\sin x}{1+\cos x}\,dx=-\int\frac{dt}{t}=-\log|t|+\mathrm{C}.\] Hence \[\int\frac{\sin x}{1+\cos x}\,dx=-\log\left|1+\cos x\right|+\mathrm{C},\qquad \cos x\neq-1.\]