SolveItClass 12 · NCERT

NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.3 11–20 (part 8 of 27)

  1. Find the integrals of the functions in Exercises $\displaystyle 1$ to $\displaystyle 22$:

    Exercise 11

    cos⁡42x\displaystyle \cos ^{4} 2 x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \frac{3 x}{8}+\frac{1}{8} \sin 4 x+\frac{1}{64} \sin 8 x+C\)
    Use \(\displaystyle \cos^{2}\theta=\dfrac{1+\cos 2\theta}{2}\) twice. With \(\displaystyle \theta=2x\), \[\cos^{4}2x=\left(\frac{1+\cos 4x}{2}\right)^{2}=\frac{1+2\cos 4x+\cos^{2}4x}{4}.\] Reduce the leftover square, \(\displaystyle \cos^{2}4x=\dfrac{1+\cos 8x}{2}\): \[\cos^{4}2x=\frac{1}{4}\left[1+2\cos 4x+\frac{1+\cos 8x}{2}\right]=\frac{3}{8}+\frac{\cos 4x}{2}+\frac{\cos 8x}{8}.\] Integrating, \[\int\cos^{4}2x\,dx=\frac{3x}{8}+\frac{1}{2}\cdot\frac{\sin 4x}{4}+\frac{1}{8}\cdot\frac{\sin 8x}{8}+\mathrm{C}.\] Final answer: \(\displaystyle \dfrac{3x}{8}+\dfrac{\sin 4x}{8}+\dfrac{\sin 8x}{64}+\mathrm{C}\).

  2. Exercise 12

    sin⁡2x1+cos⁡x\displaystyle \frac{\sin ^{2} x}{1+\cos x}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle x-\sin x+\mathrm{C}\)
    Use the Pythagorean identity \(\displaystyle \sin^{2}x=1-\cos^{2}x\) and factor the difference of squares, so the denominator cancels: \[\frac{\sin^{2}x}{1+\cos x}=\frac{1-\cos^{2}x}{1+\cos x}=\frac{(1-\cos x)(1+\cos x)}{1+\cos x}=1-\cos x,\] the cancellation being legitimate wherever \(\displaystyle 1+\cos x\neq0\). Hence \[\int\frac{\sin^{2}x}{1+\cos x}\,dx=\int(1-\cos x)\,dx=x-\sin x+\mathrm{C}.\] Final answer: \(\displaystyle x-\sin x+\mathrm{C}\).

  3. Exercise 13

    cos⁡2x−cos⁡2αcos⁡x−cos⁡α\displaystyle \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle 2(\sin x+x \cos \alpha)+\mathrm{C}\)
    Write both cosines of the double angle in the same form, \(\displaystyle \cos 2\theta=2\cos^{2}\theta-1\): \[\cos 2x-\cos 2\alpha=(2\cos^{2}x-1)-(2\cos^{2}\alpha-1)=2(\cos^{2}x-\cos^{2}\alpha).\] Factor the difference of squares so the denominator cancels: \[\frac{2(\cos x-\cos\alpha)(\cos x+\cos\alpha)}{\cos x-\cos\alpha}=2(\cos x+\cos\alpha),\qquad \cos x\neq\cos\alpha.\] Here \(\displaystyle \alpha\) is a constant, so \(\displaystyle \cos\alpha\) integrates like a number: \[\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx=2\int(\cos x+\cos\alpha)\,dx=2\sin x+2x\cos\alpha+\mathrm{C}.\] Final answer: \(\displaystyle 2(\sin x+x\cos\alpha)+\mathrm{C}\).

  4. Exercise 14

    cos⁡x−sin⁡x1+sin⁡2x\displaystyle \frac{\cos x-\sin x}{1+\sin 2 x}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle -\frac{1}{\cos x+\sin x}+C\)
    Recognise the denominator as a perfect square: since \(\displaystyle \sin^{2}x+\cos^{2}x=1\) and \(\displaystyle \sin 2x=2\sin x\cos x\), \[1+\sin 2x=\sin^{2}x+\cos^{2}x+2\sin x\cos x=(\sin x+\cos x)^{2}.\] The numerator is exactly the derivative of \(\displaystyle \sin x+\cos x\), so substitute \(\displaystyle t=\sin x+\cos x\), \(\displaystyle dt=(\cos x-\sin x)\,dx\): \[\int\frac{\cos x-\sin x}{1+\sin 2x}\,dx=\int\frac{dt}{t^{2}}=-\frac{1}{t}+\mathrm{C}.\] Final answer: \(\displaystyle -\dfrac{1}{\sin x+\cos x}+\mathrm{C}\).

  5. Exercise 15

    tan⁡32xsec⁡2x\displaystyle \tan ^{3} 2 x \sec 2 x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \frac{1}{6} \sec ^{3} 2 x-\frac{1}{2} \sec 2 x+C\)
    Keep one factor of \(\displaystyle \sec 2x\tan 2x\) aside, because that is the derivative pattern of \(\displaystyle \sec 2x\), and convert the rest with \(\displaystyle \tan^{2}\theta=\sec^{2}\theta-1\): \[\tan^{3}2x\sec 2x=\tan^{2}2x\,(\sec 2x\tan 2x)=(\sec^{2}2x-1)\,(\sec 2x\tan 2x).\] Put \(\displaystyle u=\sec 2x\); then \(\displaystyle du=2\sec 2x\tan 2x\,dx\), so \(\displaystyle \sec 2x\tan 2x\,dx=\tfrac12\,du\): \[\int\tan^{3}2x\sec 2x\,dx=\frac{1}{2}\int(u^{2}-1)\,du=\frac{1}{2}\left[\frac{u^{3}}{3}-u\right]+\mathrm{C}.\] Substituting back \(\displaystyle u=\sec 2x\): Final answer: \(\displaystyle \dfrac{\sec^{3}2x}{6}-\dfrac{\sec 2x}{2}+\mathrm{C}\).

  6. Exercise 16

    tan⁡4x\displaystyle \tan ^{4} x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \frac{1}{3} \tan ^{3} x-\tan x+x+\mathrm{C}\)
    Peel off \(\displaystyle \tan^{2}x\) using \(\displaystyle \tan^{2}x=\sec^{2}x-1\), which manufactures the \(\displaystyle \sec^{2}x\,dx\) needed for a substitution: \[\tan^{4}x=\tan^{2}x(\sec^{2}x-1)=\tan^{2}x\sec^{2}x-\tan^{2}x.\] For the first piece put \(\displaystyle t=\tan x,\;dt=\sec^{2}x\,dx\), giving \(\displaystyle \int t^{2}dt=\dfrac{\tan^{3}x}{3}\). For the second piece apply the identity once more: \[\int\tan^{2}x\,dx=\int(\sec^{2}x-1)\,dx=\tan x-x.\] Therefore \[\int\tan^{4}x\,dx=\frac{\tan^{3}x}{3}-(\tan x-x)+\mathrm{C}.\] Final answer: \(\displaystyle \dfrac{\tan^{3}x}{3}-\tan x+x+\mathrm{C}\).

  7. Exercise 17

    sin⁡3x+cos⁡3xsin⁡2xcos⁡2x\displaystyle \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \sec x-\operatorname{cosec} x+\mathrm{C}\)
    Split the fraction over the two terms of the numerator and cancel: \[\frac{\sin^{3}x+\cos^{3}x}{\sin^{2}x\cos^{2}x}=\frac{\sin^{3}x}{\sin^{2}x\cos^{2}x}+\frac{\cos^{3}x}{\sin^{2}x\cos^{2}x}=\frac{\sin x}{\cos^{2}x}+\frac{\cos x}{\sin^{2}x}.\] Each piece is a standard derivative in disguise: \[\frac{\sin x}{\cos^{2}x}=\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}=\sec x\tan x,\qquad \frac{\cos x}{\sin^{2}x}=\mathrm{cosec}\,x\cot x.\] Using \(\displaystyle \int\sec x\tan x\,dx=\sec x\) and \(\displaystyle \int\mathrm{cosec}\,x\cot x\,dx=-\,\mathrm{cosec}\,x\) (mind that minus sign): \[\int\frac{\sin^{3}x+\cos^{3}x}{\sin^{2}x\cos^{2}x}\,dx=\sec x-\mathrm{cosec}\,x+\mathrm{C}.\] Final answer: \(\displaystyle \sec x-\mathrm{cosec}\,x+\mathrm{C}\).

  8. Exercise 18

    cos⁡2x+2sin⁡2xcos⁡2x\displaystyle \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \tan x+\mathrm{C}\)
    Choose the form of \(\displaystyle \cos 2x\) that matches the other term in the numerator, namely \(\displaystyle \cos 2x=1-2\sin^{2}x\): \[\cos 2x+2\sin^{2}x=(1-2\sin^{2}x)+2\sin^{2}x=1.\] The whole numerator collapses to \(\displaystyle 1\), so \[\int\frac{\cos 2x+2\sin^{2}x}{\cos^{2}x}\,dx=\int\frac{dx}{\cos^{2}x}=\int\sec^{2}x\,dx.\] Final answer: \(\displaystyle \tan x+\mathrm{C}\).

  9. Exercise 19

    1sin⁡xcos⁡3x\displaystyle \frac{1}{\sin x \cos ^{3} x}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \log |\tan x|+\frac{1}{2} \tan ^{2} x+C\)
    Force the integrand into a function of \(\displaystyle \tan x\) by dividing numerator and denominator by \(\displaystyle \cos^{4}x\). Since \(\displaystyle \sin x\cos^{3}x=\cos^{4}x\tan x\), \[\frac{1}{\sin x\cos^{3}x}=\frac{\sec^{4}x}{\tan x}=\frac{(1+\tan^{2}x)\sec^{2}x}{\tan x},\] using \(\displaystyle \sec^{2}x=1+\tan^{2}x\) on one of the two factors of \(\displaystyle \sec^{2}x\) and keeping the other for the substitution. Put \(\displaystyle t=\tan x,\;dt=\sec^{2}x\,dx\): \[\int\frac{dx}{\sin x\cos^{3}x}=\int\frac{1+t^{2}}{t}\,dt=\int\left(\frac{1}{t}+t\right)dt=\log|t|+\frac{t^{2}}{2}+\mathrm{C}.\] Final answer: \(\displaystyle \log|\tan x|+\dfrac{\tan^{2}x}{2}+\mathrm{C}\).

  10. Exercise 20

    cos⁡2x(cos⁡x+sin⁡x)2\displaystyle \frac{\cos 2 x}{(\cos x+\sin x)^{2}}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \log |\cos x+\sin x|+\mathrm{C}\)
    Expand \(\displaystyle \cos 2x\) as a difference of squares, \(\displaystyle \cos 2x=\cos^{2}x-\sin^{2}x=(\cos x-\sin x)(\cos x+\sin x)\). Then one factor of the denominator cancels: \[\frac{\cos 2x}{(\cos x+\sin x)^{2}}=\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^{2}}=\frac{\cos x-\sin x}{\cos x+\sin x}.\] The numerator is now the derivative of the denominator, so put \(\displaystyle t=\cos x+\sin x\), \(\displaystyle dt=(\cos x-\sin x)\,dx\): \[\int\frac{\cos 2x}{(\cos x+\sin x)^{2}}\,dx=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\] Final answer: \(\displaystyle \log|\cos x+\sin x|+\mathrm{C}\).