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NCERT Solutions · Class 12 Mathematics Integrals

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Miscellaneous Exercise 1–10 (part 24 of 27)

  1. Integrate the functions in Exercises $\displaystyle 1$ to 23.

    Exercise 1

    1x−x3\displaystyle \frac{1}{x-x^{3}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \log \left|\frac{x^{2}}{1-x^{2}}\right|+\mathrm{C}\)
    Factorise the denominator and resolve into partial fractions: \[\frac{1}{x-x^{3}}=\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}. \] Clearing denominators, \(\displaystyle 1=A(1-x)(1+x)+Bx(1+x)+Cx(1-x)\). Putting \(\displaystyle x=0\) gives \(\displaystyle A=1\); \(\displaystyle x=1\) gives \(\displaystyle 1=2B\), so \(\displaystyle B=\frac12\); \(\displaystyle x=-1\) gives \(\displaystyle 1=-2C\), so \(\displaystyle C=-\frac12\). Hence \[\int\frac{dx}{x-x^{3}}=\int\frac{dx}{x}+\frac12\int\frac{dx}{1-x}-\frac12\int\frac{dx}{1+x}. \] The middle integral carries a sign from \(\displaystyle \frac{d}{dx}(1-x)=-1\), so \(\displaystyle \int\frac{dx}{1-x}=-\log|1-x|\). Therefore \[\int\frac{dx}{x-x^{3}}=\log|x|-\frac12\log|1-x|-\frac12\log|1+x|+\mathrm{C}=\log|x|-\frac12\log\left|1-x^{2}\right|+\mathrm{C}. \]

  2. Exercise 2

    1x+a+x+b\displaystyle \frac{1}{\sqrt{x+a}+\sqrt{x+b}}

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    NCERT’s answer
    \(\displaystyle \frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+\mathrm{C}\)
    Rationalise, because \(\displaystyle \left(\sqrt{x+a}\right)^{2}-\left(\sqrt{x+b}\right)^{2}=a-b\) is a constant (this needs \(\displaystyle a\neq b\)): \[\frac{1}{\sqrt{x+a}+\sqrt{x+b}}=\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}=\frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}. \] Now integrate term by term using \(\displaystyle \int (x+k)^{1/2}dx=\frac23 (x+k)^{3/2}\): \[\int\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}=\frac{1}{a-b}\left[\frac23 (x+a)^{3/2}-\frac23 (x+b)^{3/2}\right]+\mathrm{C} \] \[=\frac{2}{3(a-b)}\left[(x+a)^{3/2}-(x+b)^{3/2}\right]+\mathrm{C}. \]

  3. Exercise 3

    1xax−x2\displaystyle \frac{1}{x \sqrt{a x-x^{2}}} [Hint:Put x=at\displaystyle x=\frac{a}{t} ]

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    NCERT’s answer
    \(\displaystyle -\frac{2}{a} \sqrt{\frac{(a-x)}{x}}+\mathrm{C}\)
    Follow the hint and put \(\displaystyle x=\dfrac{a}{t}\), so \(\displaystyle dx=-\dfrac{a}{t^{2}}dt\). Then \[ax-x^{2}=\frac{a^{2}}{t}-\frac{a^{2}}{t^{2}}=\frac{a^{2}(t-1)}{t^{2}},\qquad \sqrt{ax-x^{2}}=\frac{a}{t}\sqrt{t-1}, \] where \(\displaystyle t>0\) because the integrand is only defined for \(\displaystyle 0<x<a\). Hence \[x\sqrt{ax-x^{2}}=\frac{a}{t}\cdot\frac{a}{t}\sqrt{t-1}=\frac{a^{2}}{t^{2}}\sqrt{t-1}, \] and the integral becomes \[\int\frac{dx}{x\sqrt{ax-x^{2}}}=\int\frac{-\dfrac{a}{t^{2}}\,dt}{\dfrac{a^{2}}{t^{2}}\sqrt{t-1}}=-\frac{1}{a}\int\frac{dt}{\sqrt{t-1}}=-\frac{2}{a}\sqrt{t-1}+\mathrm{C}. \] Back-substituting \(\displaystyle t=\dfrac{a}{x}\), \[=-\frac{2}{a}\sqrt{\frac{a}{x}-1}+\mathrm{C}=-\frac{2}{a}\sqrt{\frac{a-x}{x}}+\mathrm{C}. \]

  4. Exercise 4

    1x2(x4+1)34\displaystyle \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}

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    NCERT’s answer
    \(\displaystyle -\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+\mathrm{C}\)
    The step that makes this work is taking \(\displaystyle x^{4}\) out of the bracket before substituting (take \(\displaystyle x>0\)): \[\left(x^{4}+1\right)^{3/4}=\left[x^{4}\left(1+x^{-4}\right)\right]^{3/4}=x^{3}\left(1+\frac{1}{x^{4}}\right)^{3/4}. \] So the integrand is \[\frac{1}{x^{2}\left(x^{4}+1\right)^{3/4}}=\frac{1}{x^{5}}\left(1+\frac{1}{x^{4}}\right)^{-3/4}. \] Put \(\displaystyle t=1+\dfrac{1}{x^{4}}\), so \(\displaystyle dt=-\dfrac{4}{x^{5}}dx\), i.e. \(\displaystyle \dfrac{dx}{x^{5}}=-\dfrac{dt}{4}\): \[\int\frac{dx}{x^{2}\left(x^{4}+1\right)^{3/4}}=-\frac14\int t^{-3/4}dt=-\frac14\cdot 4\,t^{1/4}+\mathrm{C}=-t^{1/4}+\mathrm{C}. \] \[=-\left(1+\frac{1}{x^{4}}\right)^{1/4}+\mathrm{C}. \]

  5. Exercise 5

    1x12+x13\displaystyle \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} [Hint: 1x12+x13=1x13(1+x16)\displaystyle \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}, put x=t6\displaystyle x=t^{6} ]

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    NCERT’s answer
    \(\displaystyle 2 \sqrt{x}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log \left(1+x^{\frac{1}{6}}\right)+\mathrm{C}\)
    Both fractional powers are cleared at once by \(\displaystyle x=t^{6}\), since \(\displaystyle 6\) is the l.c.m. of \(\displaystyle 2\) and \(\displaystyle 3\). Then \(\displaystyle dx=6t^{5}dt\), \(\displaystyle x^{1/2}=t^{3}\), \(\displaystyle x^{1/3}=t^{2}\): \[\int\frac{dx}{x^{1/2}+x^{1/3}}=\int\frac{6t^{5}}{t^{3}+t^{2}}dt=\int\frac{6t^{5}}{t^{2}(t+1)}dt=6\int\frac{t^{3}}{t+1}dt. \] This rational function is improper, so divide before integrating: \(\displaystyle \dfrac{t^{3}}{t+1}=t^{2}-t+1-\dfrac{1}{t+1}\). Hence \[6\int\left(t^{2}-t+1-\frac{1}{t+1}\right)dt=2t^{3}-3t^{2}+6t-6\log|t+1|+\mathrm{C}. \] With \(\displaystyle t=x^{1/6}\) (so \(\displaystyle t>0\)), \[=2\sqrt{x}-3x^{1/3}+6x^{1/6}-6\log\left(1+x^{1/6}\right)+\mathrm{C}. \]

  6. Exercise 6

    5x(x+1)(x2+9)\displaystyle \frac{5 x}{(x+1)\left(x^{2}+9\right)}

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    NCERT’s answer
    \(\displaystyle -\frac{1}{2} \log |x+1|+\frac{1}{4} \log \left(x^{2}+9\right)+\frac{3}{2} \tan ^{-1} \frac{x}{3}+\mathrm{C}\)
    The denominator has a linear factor and an irreducible quadratic, so the partial-fraction form must be \[\frac{5x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+9}, \] giving \(\displaystyle 5x=A\left(x^{2}+9\right)+(Bx+C)(x+1)\). Putting \(\displaystyle x=-1\): \(\displaystyle -5=10A\), so \(\displaystyle A=-\frac12\). Comparing coefficients of \(\displaystyle x^{2}\): \(\displaystyle 0=A+B\), so \(\displaystyle B=\frac12\). Comparing constants: \(\displaystyle 0=9A+C\), so \(\displaystyle C=\frac92\). \[\int\frac{5x\,dx}{(x+1)\left(x^{2}+9\right)}=-\frac12\int\frac{dx}{x+1}+\frac12\int\frac{x\,dx}{x^{2}+9}+\frac92\int\frac{dx}{x^{2}+9}. \] Using \(\displaystyle \int\frac{x\,dx}{x^{2}+9}=\frac12\log\left(x^{2}+9\right)\) and \(\displaystyle \int\frac{dx}{x^{2}+9}=\frac13\tan^{-1}\frac{x}{3}\), \[=-\frac12\log|x+1|+\frac14\log\left(x^{2}+9\right)+\frac32\tan^{-1}\frac{x}{3}+\mathrm{C}. \]

  7. Exercise 7

    sin⁡xsin⁡(x−a)\displaystyle \frac{\sin x}{\sin (x-a)}

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    NCERT’s answer
    \(\displaystyle \sin a \log |\sin (x-a)|+x \cos a+\mathrm{C}\)
    Numerator and denominator have different angles, so rewrite the numerator's angle in terms of the denominator's, \(\displaystyle x=(x-a)+a\), and expand by the compound-angle formula: \[\sin x=\sin\big((x-a)+a\big)=\sin (x-a)\cos a+\cos (x-a)\sin a. \] Dividing by \(\displaystyle \sin (x-a)\), \[\frac{\sin x}{\sin (x-a)}=\cos a+\sin a\,\cot (x-a). \] Here \(\displaystyle a\) is a constant, so \[\int\frac{\sin x}{\sin (x-a)}dx=\cos a\int dx+\sin a\int\cot (x-a)\,dx=x\cos a+\sin a\,\log|\sin (x-a)|+\mathrm{C}. \]

  8. Exercise 8

    e5log⁡x−e4log⁡xe3log⁡x−e2log⁡x\displaystyle \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}

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    NCERT’s answer
    \(\displaystyle \frac{x^{3}}{3}+C\)
    Simplify before integrating, using \(\displaystyle e^{n\log x}=e^{\log x^{n}}=x^{n}\), valid for \(\displaystyle x>0\): \[\frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}=\frac{x^{5}-x^{4}}{x^{3}-x^{2}}=\frac{x^{4}(x-1)}{x^{2}(x-1)}=x^{2},\qquad x>0,\;x\neq 1. \] Hence \[\int\frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}\,dx=\int x^{2}dx=\frac{x^{3}}{3}+\mathrm{C}. \]

  9. Exercise 9

    cos⁡x4−sin⁡2x\displaystyle \frac{\cos x}{\sqrt{4-\sin ^{2} x}}

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    NCERT’s answer
    \(\displaystyle \sin ^{-1}\left(\frac{\sin x}{2}\right)+\mathrm{C}\)
    The numerator is exactly the derivative of \(\displaystyle \sin x\), so substitute \(\displaystyle t=\sin x\), \(\displaystyle dt=\cos x\,dx\): \[\int\frac{\cos x}{\sqrt{4-\sin^{2}x}}dx=\int\frac{dt}{\sqrt{2^{2}-t^{2}}}. \] By the standard form \(\displaystyle \int\frac{dt}{\sqrt{a^{2}-t^{2}}}=\sin^{-1}\frac{t}{a}+\mathrm{C}\), \[=\sin^{-1}\frac{t}{2}+\mathrm{C}=\sin^{-1}\left(\frac{\sin x}{2}\right)+\mathrm{C}. \]

  10. Exercise 10

    sin⁡8−cos⁡8x1−2sin⁡2xcos⁡2x\displaystyle \frac{\sin ^{8}-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}

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    NCERT’s answer
    \(\displaystyle -\frac{1}{2} \sin 2 x+C\)
    Factorise the numerator as a difference of squares twice, using \(\displaystyle \sin^{2}x+\cos^{2}x=1\): \[\sin^{8}x-\cos^{8}x=\left(\sin^{4}x-\cos^{4}x\right)\left(\sin^{4}x+\cos^{4}x\right)=\left(\sin^{2}x-\cos^{2}x\right)\left(\sin^{4}x+\cos^{4}x\right). \] The denominator is precisely that second factor: \[1-2\sin^{2}x\cos^{2}x=\left(\sin^{2}x+\cos^{2}x\right)^{2}-2\sin^{2}x\cos^{2}x=\sin^{4}x+\cos^{4}x. \] So the integrand reduces to \(\displaystyle \sin^{2}x-\cos^{2}x=-\cos 2x\), and \[\int\frac{\sin^{8}x-\cos^{8}x}{1-2\sin^{2}x\cos^{2}x}dx=-\int\cos 2x\,dx=-\frac{\sin 2x}{2}+\mathrm{C}. \]