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NCERT Exemplar · Class 12 Mathematics Integrals

63 questions · 63 still being checked

EXERCISE 7.3 11–20 (part 2 of 6)

  1. Evaluate the following:

    Exercise 11

    ∫a+xa−x\displaystyle \int \sqrt{\frac{a+x}{a-x}}

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    NCERT’s answer
    \(\displaystyle -a\left[\cos ^{-1}\left(\frac{x}{a}\right)+\sqrt{1-\frac{x^2}{a^2}}\right]+c\)
    Rationalise the integrand (\(\displaystyle a>0\), \(\displaystyle |x|<a\)).\[\sqrt{\frac{a+x}{a-x}}=\frac{a+x}{\sqrt{(a-x)(a+x)}}=\frac{a+x}{\sqrt{a^2-x^2}} \]\[\int\sqrt{\frac{a+x}{a-x}}\,dx=a\int\frac{dx}{\sqrt{a^2-x^2}}+\int\frac{x\,dx}{\sqrt{a^2-x^2}} \]For the second integral put \(\displaystyle a^2-x^2=u\).\[x\,dx=-\tfrac{1}{2}\,du \]\[\int\frac{x\,dx}{\sqrt{a^2-x^2}}=-\frac{1}{2}\int u^{-1/2}\,du=-\sqrt{a^2-x^2} \]\[\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a} \]Answer: \(\displaystyle a\sin^{-1}\dfrac{x}{a}-\sqrt{a^2-x^2}+C\)

  2. Exercise 12

    ∫x121+x34dx\displaystyle \int \frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}} d x (Hint : Put x=z4\displaystyle x=z^4)

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    NCERT’s answer
    \(\displaystyle \frac{4}{3}\left[x^{3 / 4}-\log \left|1+x^{\frac{3}{4}}\right|\right]+c\)
    Put \(\displaystyle x=z^4\).\[dx=4z^3\,dz,\quad x^{1/2}=z^2,\quad x^{3/4}=z^3 \]\[\int\frac{x^{1/2}}{1+x^{3/4}}\,dx=4\int\frac{z^5}{1+z^3}\,dz \]\[z^5=z^2(1+z^3)-z^2 \]\[4\int\left(z^2-\frac{z^2}{1+z^3}\right)dz=4\left[\frac{z^3}{3}-\frac{1}{3}\ln\left|1+z^3\right|\right]+C \]\[=\frac{4}{3}\left[x^{3/4}-\ln\left(1+x^{3/4}\right)\right]+C \]Answer: \(\displaystyle \dfrac{4}{3}\left[x^{3/4}-\ln\left(1+x^{3/4}\right)\right]+C\)

  3. Exercise 13

    ∫1+x2x4dx\displaystyle \int \frac{\sqrt{1+x^2}}{x^4} d x

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    NCERT’s answer
    \(\displaystyle \frac{-1}{3}\left(1+\frac{1}{x^2}\right)^{\frac{3}{2}}+c\)
    Put \(\displaystyle x=\tan\theta\).\[dx=\sec^2\theta\,d\theta,\quad \sqrt{1+x^2}=\sec\theta \]\[\int\frac{\sqrt{1+x^2}}{x^4}\,dx=\int\frac{\sec^3\theta}{\tan^4\theta}\,d\theta=\int\frac{\cos\theta}{\sin^4\theta}\,d\theta \]Put \(\displaystyle \sin\theta=s\).\[\cos\theta\,d\theta=ds \]\[\int s^{-4}\,ds=-\frac{1}{3s^3}+C \]\[s=\sin\theta=\frac{\tan\theta}{\sec\theta}=\frac{x}{\sqrt{1+x^2}} \]\[-\frac{1}{3s^3}=-\frac{\left(1+x^2\right)^{3/2}}{3x^3} \]Answer: \(\displaystyle -\dfrac{\left(1+x^2\right)^{3/2}}{3x^3}+C\)

  4. Exercise 14

    ∫dx16−9x2\displaystyle \int \frac{d x}{\sqrt{16-9 x^2}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{3} \sin ^{-1} \frac{3 x}{4}+c\)
    Take $\displaystyle 9$ out of the root.\[16-9x^2=9\left[\left(\tfrac{4}{3}\right)^2-x^2\right] \]\[\int\frac{dx}{\sqrt{16-9x^2}}=\frac{1}{3}\int\frac{dx}{\sqrt{\left(\dfrac{4}{3}\right)^2-x^2}} \]\[\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a},\quad a=\tfrac{4}{3} \]\[=\frac{1}{3}\sin^{-1}\frac{3x}{4}+C \]Answer: \(\displaystyle \dfrac{1}{3}\sin^{-1}\dfrac{3x}{4}+C\)

  5. Exercise 15

    ∫dt3t−2t2\displaystyle \int \frac{d t}{\sqrt{3 t-2 t^2}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4 t-3}{3}\right)+c\)
    Complete the square under the root.\[3t-2t^2=-2\left(t^2-\tfrac{3}{2}t\right)=2\left[\left(\tfrac{3}{4}\right)^2-\left(t-\tfrac{3}{4}\right)^2\right] \]\[\int\frac{dt}{\sqrt{3t-2t^2}}=\frac{1}{\sqrt{2}}\int\frac{dt}{\sqrt{\left(\dfrac{3}{4}\right)^2-\left(t-\dfrac{3}{4}\right)^2}} \]\[=\frac{1}{\sqrt{2}}\sin^{-1}\frac{t-\dfrac{3}{4}}{\dfrac{3}{4}}+C \]\[=\frac{1}{\sqrt{2}}\sin^{-1}\frac{4t-3}{3}+C \]Answer: \(\displaystyle \dfrac{1}{\sqrt{2}}\sin^{-1}\dfrac{4t-3}{3}+C\)

  6. Exercise 16

    ∫3x−1x2+9dx\displaystyle \int \frac{3 x-1}{\sqrt{x^2+9}} d x

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    NCERT’s answer
    \(\displaystyle 3 \sqrt{x^2+9}-\log \left|x+\sqrt{x^2+9}\right|+c\)
    Split the numerator.\[\int\frac{3x-1}{\sqrt{x^2+9}}\,dx=3\int\frac{x\,dx}{\sqrt{x^2+9}}-\int\frac{dx}{\sqrt{x^2+9}} \]For the first integral put \(\displaystyle x^2+9=u\).\[x\,dx=\tfrac{1}{2}\,du \]\[3\int\frac{x\,dx}{\sqrt{x^2+9}}=\frac{3}{2}\int u^{-1/2}\,du=3\sqrt{x^2+9} \]\[\int\frac{dx}{\sqrt{x^2+a^2}}=\ln\left|x+\sqrt{x^2+a^2}\right|,\quad a=3 \]Answer: \(\displaystyle 3\sqrt{x^2+9}-\ln\left|x+\sqrt{x^2+9}\right|+C\)

  7. Exercise 17

    ∫5−2x+x2dx\displaystyle \int \sqrt{5-2 x+x^2} d x

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    NCERT’s answer
    \(\displaystyle \frac{x-1}{2} \sqrt{5-2 x+x^2}+2 \log \left|x-1+\sqrt{5-2 x+x^2}\right|+c\)
    \[x^2-2x+5=(x-1)^2+2^2 \] Put \(\displaystyle u=x-1\), so \(\displaystyle du=dx\), and use the standard form: \[\int\sqrt{u^2+a^2}\,du=\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln\left|u+\sqrt{u^2+a^2}\right|+C \] \[\int\sqrt{(x-1)^2+2^2}\,dx=\frac{x-1}{2}\sqrt{(x-1)^2+4}+\frac{4}{2}\ln\left|(x-1)+\sqrt{(x-1)^2+4}\right|+C \] Answer: \(\displaystyle \dfrac{x-1}{2}\sqrt{x^2-2x+5}+2\ln\left|x-1+\sqrt{x^2-2x+5}\right|+C\)

  8. Exercise 18

    ∫xx4−1dx\displaystyle \int \frac{x}{x^4-1} d x

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    NCERT’s answer
    \(\displaystyle \frac{1}{4}\left\{\log \left|x^2-1\right|-\log \left|x^2+1\right|\right\}+c\)
    Put \(\displaystyle x^2=t\), so \(\displaystyle 2x\,dx=dt\). \[\int\frac{x\,dx}{x^4-1}=\frac12\int\frac{dt}{t^2-1} \] \[\int\frac{dt}{t^2-1}=\frac12\ln\left|\frac{t-1}{t+1}\right| \] \[\int\frac{x\,dx}{x^4-1}=\frac14\ln\left|\frac{x^2-1}{x^2+1}\right|+C \] Answer: \(\displaystyle \dfrac14\ln\left|\dfrac{x^2-1}{x^2+1}\right|+C\)

  9. Exercise 19

    ∫x21−x4dx\displaystyle \int \frac{x^2}{1-x^4} d x put x2=t\displaystyle x^2=t

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    NCERT’s answer
    \(\displaystyle \frac{1}{4}\left\{\log \left|\frac{1+x}{1-x}\right|\right\}-\frac{1}{2} \tan ^{-1} x+c\)
    With \(\displaystyle x^2=t\), the integrand is split by partial fractions: \[\frac{x^2}{1-x^4}=\frac{t}{1-t^2}=\frac12\left[\frac{1}{1-t}-\frac{1}{1+t}\right] \] \[\frac{x^2}{1-x^4}=\frac12\left[\frac{1}{1-x^2}-\frac{1}{1+x^2}\right] \] \[\int\frac{dx}{1-x^2}=\frac12\ln\left|\frac{1+x}{1-x}\right|,\qquad \int\frac{dx}{1+x^2}=\tan^{-1}x \] \[\int\frac{x^2}{1-x^4}\,dx=\frac14\ln\left|\frac{1+x}{1-x}\right|-\frac12\tan^{-1}x+C \] Answer: \(\displaystyle \dfrac14\ln\left|\dfrac{1+x}{1-x}\right|-\dfrac12\tan^{-1}x+C\)

  10. Exercise 20

    ∫2ax−x2dx\displaystyle \int \sqrt{2 a x-x^2} d x

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    NCERT’s answer
    \(\displaystyle \frac{x-a}{2} \sqrt{2 a x-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)+c\)
    \[2ax-x^2=a^2-(x-a)^2 \] Put \(\displaystyle u=x-a\), so \(\displaystyle du=dx\), and use the standard form: \[\int\sqrt{a^2-u^2}\,du=\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}\frac{u}{a}+C \] \[\int\sqrt{2ax-x^2}\,dx=\frac{x-a}{2}\sqrt{2ax-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x-a}{a}+C \] Answer: \(\displaystyle \dfrac{x-a}{2}\sqrt{2ax-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x-a}{a}+C\)