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NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.3 1–10 (part 7 of 27)

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

    Exercise 1

    sin⁡2(2x+5)\displaystyle \sin ^{2}(2 x+5)

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    NCERT’s answer
    \(\displaystyle \frac{x}{2}-\frac{1}{8} \sin (4 x+10)+\mathrm{C}\)
    Apply the power-reducing identity \(\displaystyle \sin^{2}\theta=\dfrac{1-\cos 2\theta}{2}\) with \(\displaystyle \theta=2x+5\), so that \(\displaystyle 2\theta=4x+10\): \[\sin^{2}(2x+5)=\frac{1-\cos(4x+10)}{2}.\] Now integrate, using \(\displaystyle \int\cos(ax+b)\,dx=\dfrac{\sin(ax+b)}{a}\) (the factor \(\displaystyle \tfrac1a\) is what the chain rule forces): \[\int\sin^{2}(2x+5)\,dx=\frac{1}{2}\int dx-\frac{1}{2}\int\cos(4x+10)\,dx=\frac{x}{2}-\frac{1}{2}\cdot\frac{\sin(4x+10)}{4}+\mathrm{C}.\] Final answer: \(\displaystyle \dfrac{x}{2}-\dfrac{1}{8}\sin(4x+10)+\mathrm{C}\).

  2. Exercise 2

    sin⁡3xcos⁡4x\displaystyle \sin 3 x \cos 4 x

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    NCERT’s answer
    \(\displaystyle -\frac{1}{14} \cos 7 x+\frac{1}{2} \cos x+\mathrm{C}\)
    A product of sines and cosines has no direct antiderivative, so convert it to a sum by \(\displaystyle \sin A\cos B=\tfrac12\big[\sin(A+B)+\sin(A-B)\big]\). With \(\displaystyle A=3x,\;B=4x\): \[\sin 3x\cos 4x=\tfrac12\big[\sin 7x+\sin(-x)\big]=\tfrac12\big[\sin 7x-\sin x\big],\] using \(\displaystyle \sin(-x)=-\sin x\). Integrating, \[\int\sin 3x\cos 4x\,dx=\frac{1}{2}\left[-\frac{\cos 7x}{7}+\cos x\right]+\mathrm{C}.\] Final answer: \(\displaystyle -\dfrac{\cos 7x}{14}+\dfrac{\cos x}{2}+\mathrm{C}\).

  3. Exercise 3

    cos⁡2xcos⁡4xcos⁡6x\displaystyle \cos 2 x \cos 4 x \cos 6 x

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    NCERT’s answer
    \(\displaystyle \frac{1}{4}\left[\frac{1}{12} \sin 12 x+x+\frac{1}{8} \sin 8 x+\frac{1}{4} \sin 4 x\right]+C\)
    Reduce the triple product to sums in two stages, using \(\displaystyle \cos A\cos B=\tfrac12\big[\cos(A+B)+\cos(A-B)\big]\). First pair the first two factors: \[\cos 2x\cos 4x=\tfrac12\big[\cos 6x+\cos 2x\big].\] Multiply by \(\displaystyle \cos 6x\): \[\cos 2x\cos 4x\cos 6x=\tfrac12\big[\cos^{2}6x+\cos 2x\cos 6x\big]=\tfrac12\left[\frac{1+\cos 12x}{2}+\frac{\cos 8x+\cos 4x}{2}\right],\] where \(\displaystyle \cos^{2}6x=\dfrac{1+\cos 12x}{2}\) and \(\displaystyle \cos 2x\cos 6x=\tfrac12[\cos 8x+\cos 4x]\). Hence the integrand is \[\tfrac14\big[1+\cos 12x+\cos 8x+\cos 4x\big].\] Integrating term by term, \[\int\cos 2x\cos 4x\cos 6x\,dx=\frac{1}{4}\left[x+\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}\right]+\mathrm{C}.\] Final answer: \(\displaystyle \dfrac{x}{4}+\dfrac{\sin 12x}{48}+\dfrac{\sin 8x}{32}+\dfrac{\sin 4x}{16}+\mathrm{C}\).

  4. Exercise 4

    sin⁡3(2x+1)\displaystyle \sin ^{3}(2 x+1)

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    NCERT’s answer
    \(\displaystyle -\frac{1}{2} \cos (2 x+1)+\frac{1}{6} \cos ^{3}(2 x+1)+\mathrm{C}\)
    Use the triple-angle identity \(\displaystyle \sin 3\theta=3\sin\theta-4\sin^{3}\theta\), i.e. \(\displaystyle \sin^{3}\theta=\dfrac{3\sin\theta-\sin 3\theta}{4}\). Take \(\displaystyle \theta=2x+1\), so \(\displaystyle 3\theta=6x+3\): \[\sin^{3}(2x+1)=\frac{3\sin(2x+1)-\sin(6x+3)}{4}.\] Integrate, with \(\displaystyle \int\sin(ax+b)\,dx=-\dfrac{\cos(ax+b)}{a}\): \[\int\sin^{3}(2x+1)\,dx=\frac{3}{4}\left(-\frac{\cos(2x+1)}{2}\right)-\frac{1}{4}\left(-\frac{\cos(6x+3)}{6}\right)+\mathrm{C}.\] Final answer: \(\displaystyle -\dfrac{3}{8}\cos(2x+1)+\dfrac{1}{24}\cos(6x+3)+\mathrm{C}\). (Equivalently, splitting off one sine and putting \(\displaystyle t=\cos(2x+1)\) gives \(\displaystyle -\tfrac12\cos(2x+1)+\tfrac16\cos^{3}(2x+1)+\mathrm{C}\); the two answers differ only by a constant.)

  5. Exercise 5

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

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    NCERT’s answer
    \(\displaystyle \frac{1}{6} \cos ^{6} x-\frac{1}{4} \cos ^{4} x+\mathrm{C}\)
    Pull the product into a single angle with \(\displaystyle \sin x\cos x=\tfrac12\sin 2x\): \[\sin^{3}x\cos^{3}x=(\sin x\cos x)^{3}=\left(\frac{\sin 2x}{2}\right)^{3}=\frac{\sin^{3}2x}{8}.\] Now use \(\displaystyle \sin^{3}\theta=\dfrac{3\sin\theta-\sin 3\theta}{4}\) with \(\displaystyle \theta=2x\): \[\frac{\sin^{3}2x}{8}=\frac{3\sin 2x-\sin 6x}{32}.\] Integrating, \[\int\sin^{3}x\cos^{3}x\,dx=\frac{1}{32}\left[3\left(-\frac{\cos 2x}{2}\right)+\frac{\cos 6x}{6}\right]+\mathrm{C}.\] Final answer: \(\displaystyle -\dfrac{3\cos 2x}{64}+\dfrac{\cos 6x}{192}+\mathrm{C}\). (The substitution \(\displaystyle t=\cos x\) gives the equivalent form \(\displaystyle -\dfrac{\cos^{4}x}{4}+\dfrac{\cos^{6}x}{6}+\mathrm{C}\).)

  6. Exercise 6

    sin⁡xsin⁡2xsin⁡3x\displaystyle \sin x \sin 2 x \sin 3 x

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    NCERT’s answer
    \(\displaystyle \frac{1}{4}\left[\frac{1}{6} \cos 6 x-\frac{1}{4} \cos 4 x-\frac{1}{2} \cos 2 x\right]+\mathrm{C}\)
    Pair the two factors whose angles combine cleanly. By \(\displaystyle \sin A\sin B=\tfrac12\big[\cos(A-B)-\cos(A+B)\big]\), \[\sin x\sin 3x=\tfrac12\big[\cos 2x-\cos 4x\big].\] Multiply by \(\displaystyle \sin 2x\) and use \(\displaystyle \sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]\): \[\sin x\sin 2x\sin 3x=\tfrac12\big[\sin 2x\cos 2x-\sin 2x\cos 4x\big]=\tfrac12\left[\frac{\sin 4x}{2}-\frac{\sin 6x-\sin 2x}{2}\right],\] since \(\displaystyle \sin 2x\cos 4x=\tfrac12[\sin 6x+\sin(-2x)]=\tfrac12[\sin 6x-\sin 2x]\). So the integrand is \[\tfrac14\big[\sin 4x-\sin 6x+\sin 2x\big].\] Integrating, \[\int\sin x\sin 2x\sin 3x\,dx=\frac{1}{4}\left[-\frac{\cos 4x}{4}+\frac{\cos 6x}{6}-\frac{\cos 2x}{2}\right]+\mathrm{C}.\] Final answer: \(\displaystyle -\dfrac{\cos 4x}{16}+\dfrac{\cos 6x}{24}-\dfrac{\cos 2x}{8}+\mathrm{C}\).

  7. Exercise 7

    sin⁡4xsin⁡8x\displaystyle \sin 4 x \sin 8 x

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\left[\frac{1}{4} \sin 4 x-\frac{1}{12} \sin 12 x\right]+C\)
    Convert the product to a sum by \(\displaystyle \sin A\sin B=\tfrac12\big[\cos(A-B)-\cos(A+B)\big]\), with \(\displaystyle A=4x,\;B=8x\): \[\sin 4x\sin 8x=\tfrac12\big[\cos(-4x)-\cos 12x\big]=\tfrac12\big[\cos 4x-\cos 12x\big],\] using \(\displaystyle \cos(-\theta)=\cos\theta\). Integrating, \[\int\sin 4x\sin 8x\,dx=\frac{1}{2}\left[\frac{\sin 4x}{4}-\frac{\sin 12x}{12}\right]+\mathrm{C}.\] Final answer: \(\displaystyle \dfrac{\sin 4x}{8}-\dfrac{\sin 12x}{24}+\mathrm{C}\).

  8. Exercise 8

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

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    NCERT’s answer
    \(\displaystyle 2 \tan \frac{x}{2}-x+C\)
    Write numerator and denominator in half-angle form, using \(\displaystyle 1-\cos x=2\sin^{2}\dfrac{x}{2}\) and \(\displaystyle 1+\cos x=2\cos^{2}\dfrac{x}{2}\): \[\frac{1-\cos x}{1+\cos x}=\frac{2\sin^{2}(x/2)}{2\cos^{2}(x/2)}=\tan^{2}\frac{x}{2}.\] A squared tangent is never integrated directly; convert it with \(\displaystyle \tan^{2}\theta=\sec^{2}\theta-1\): \[\int\tan^{2}\frac{x}{2}\,dx=\int\left(\sec^{2}\frac{x}{2}-1\right)dx=\frac{\tan(x/2)}{1/2}-x+\mathrm{C}.\] Final answer: \(\displaystyle 2\tan\dfrac{x}{2}-x+\mathrm{C}\), valid wherever \(\displaystyle \cos x\neq-1\).

  9. Exercise 9

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

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    NCERT’s answer
    \(\displaystyle x-\tan \frac{x}{2}+C\)
    Split the numerator so the fraction becomes a constant minus a standard piece: \[\frac{\cos x}{1+\cos x}=\frac{(1+\cos x)-1}{1+\cos x}=1-\frac{1}{1+\cos x}.\] Now use the half-angle identity \(\displaystyle 1+\cos x=2\cos^{2}\dfrac{x}{2}\): \[1-\frac{1}{2\cos^{2}(x/2)}=1-\frac{1}{2}\sec^{2}\frac{x}{2}.\] Integrating, and remembering \(\displaystyle \int\sec^{2}\dfrac{x}{2}\,dx=2\tan\dfrac{x}{2}\): \[\int\frac{\cos x}{1+\cos x}\,dx=x-\frac{1}{2}\cdot 2\tan\frac{x}{2}+\mathrm{C}.\] Final answer: \(\displaystyle x-\tan\dfrac{x}{2}+\mathrm{C}\), valid wherever \(\displaystyle \cos x\neq-1\).

  10. Exercise 10

    sin⁡4x\displaystyle \sin ^{4} x

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    NCERT’s answer
    \(\displaystyle \frac{3 x}{8}-\frac{1}{4} \sin 2 x+\frac{1}{32} \sin 4 x+C\)
    Reduce the fourth power in two steps with \(\displaystyle \sin^{2}\theta=\dfrac{1-\cos 2\theta}{2}\). First, \[\sin^{4}x=\left(\frac{1-\cos 2x}{2}\right)^{2}=\frac{1-2\cos 2x+\cos^{2}2x}{4}.\] The step students skip: \(\displaystyle \cos^{2}2x\) is still a square, so reduce it too, by \(\displaystyle \cos^{2}2x=\dfrac{1+\cos 4x}{2}\): \[\sin^{4}x=\frac{1}{4}\left[1-2\cos 2x+\frac{1+\cos 4x}{2}\right]=\frac{3}{8}-\frac{\cos 2x}{2}+\frac{\cos 4x}{8}.\] Integrating, \[\int\sin^{4}x\,dx=\frac{3x}{8}-\frac{1}{2}\cdot\frac{\sin 2x}{2}+\frac{1}{8}\cdot\frac{\sin 4x}{4}+\mathrm{C}.\] Final answer: \(\displaystyle \dfrac{3x}{8}-\dfrac{\sin 2x}{4}+\dfrac{\sin 4x}{32}+\mathrm{C}\).