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NCERT Exemplar · Class 12 Mathematics Integrals

63 questions · 63 still being checked

EXERCISE 7.3 31–40 (part 4 of 6)

  1. Evaluate the following:

    Exercise 31

    ∫12dx(x−1)(2−x)\displaystyle \int_1^2 \frac{d x}{\sqrt{(x-1)(2-x)}}

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    NCERT’s answer
    \(\displaystyle \pi\)
    Complete the square.\[(x-1)(2-x) = -x^2+3x-2 = \left(\frac12\right)^2-\left(x-\frac32\right)^2 \]\[I = \int_1^2\frac{dx}{\sqrt{\left(\frac12\right)^2-\left(x-\frac32\right)^2}} \]\[= \left[\sin^{-1}\frac{x-\frac32}{\frac12}\right]_1^2 = \Big[\sin^{-1}(2x-3)\Big]_1^2 \]\[= \sin^{-1}(1)-\sin^{-1}(-1) \]\[= \frac\pi2+\frac\pi2 = \pi \]Answer: \(\displaystyle \pi\)

  2. Exercise 32

    ∫01xdx1+x2\displaystyle \int_0^1 \frac{x d x}{\sqrt{1+x^2}}

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    NCERT’s answer
    \(\displaystyle \sqrt{2}-1\)
    Put \(\displaystyle t=1+x^2\), so \(\displaystyle 2x\,dx=dt\); \(\displaystyle x:0\to1\) gives \(\displaystyle t:1\to2\).\[I = \frac12\int_1^2 t^{-1/2}\,dt \]\[= \frac12\Big[2\sqrt t\Big]_1^2 = \sqrt2-1 \]Answer: \(\displaystyle \sqrt2-1\)

  3. Exercise 33

    ∫0πxsin⁡xcos⁡2xdx\displaystyle \int_0^\pi x \sin x \cos ^2 x d x

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{3}\)
    Use \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\); \(\displaystyle \sin(\pi-x)=\sin x\) and \(\displaystyle \cos^2(\pi-x)=\cos^2 x\). \[I=\int_0^\pi x\sin x\cos^2x\,dx=\int_0^\pi (\pi-x)\sin x\cos^2x\,dx \] Adding the two forms: \[2I=\pi\int_0^\pi \sin x\cos^2x\,dx \] \[2I=\pi\left[-\frac{\cos^3x}{3}\right]_0^\pi=\pi\left(\frac13+\frac13\right)=\frac{2\pi}{3} \] \[I=\frac{\pi}{3} \] Answer: \(\displaystyle \dfrac{\pi}{3}\)

  4. Exercise 34

    ∫012dx(1+x2)1−x2\displaystyle \int_0^{\frac{1}{2}} \frac{d x}{\left(1+x^2\right) \sqrt{1-x^2}} (Hint: let x=sin⁡θ\displaystyle x=\sin \theta)

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    Put \(\displaystyle x=\sin\theta\). \[dx=\cos\theta\,d\theta,\quad \sqrt{1-x^2}=\cos\theta,\quad x:0\to\tfrac12 \;\Rightarrow\; \theta:0\to\tfrac{\pi}{6} \] \[I=\int_0^{\pi/6}\frac{d\theta}{1+\sin^2\theta} \] Divide top and bottom by \(\displaystyle \cos^2\theta\): \[I=\int_0^{\pi/6}\frac{\sec^2\theta\,d\theta}{\sec^2\theta+\tan^2\theta}=\int_0^{\pi/6}\frac{\sec^2\theta\,d\theta}{1+2\tan^2\theta} \] Put \(\displaystyle t=\tan\theta\): \[dt=\sec^2\theta\,d\theta,\quad t:0\to\tfrac{1}{\sqrt3} \] \[I=\int_0^{1/\sqrt3}\frac{dt}{1+2t^2}=\frac{1}{\sqrt2}\Big[\tan^{-1}(\sqrt2\,t)\Big]_0^{1/\sqrt3} \] \[I=\frac{1}{\sqrt2}\tan^{-1}\sqrt{\frac23} \] Answer: \(\displaystyle \dfrac{1}{\sqrt2}\tan^{-1}\sqrt{\dfrac23}\)

  5. Exercise 35

    ∫x2dxx4−x2−12\displaystyle \int \frac{x^2 d x}{x^4-x^2-12}

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    NCERT’s answer
    \(\displaystyle \frac{1}{7} \log \left|\frac{x-2}{x+2}\right|+\frac{\sqrt{3}}{7} \tan ^{-1} \frac{x}{\sqrt{3}}+c\)
    Factorise the denominator, then split into partial fractions in \(\displaystyle x^2\). \[x^4-x^2-12=(x^2-4)(x^2+3) \] \[\frac{x^2}{(x^2-4)(x^2+3)}=\frac{A}{x^2-4}+\frac{B}{x^2+3} \] \[x^2=A(x^2+3)+B(x^2-4) \] \[x^2=4:\; 4=7A \Rightarrow A=\tfrac47 \qquad x^2=-3:\; -3=-7B \Rightarrow B=\tfrac37 \] \[I=\frac47\int\frac{dx}{x^2-2^2}+\frac37\int\frac{dx}{x^2+(\sqrt3)^2} \] \[I=\frac47\cdot\frac14\ln\left|\frac{x-2}{x+2}\right|+\frac37\cdot\frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}+C \] Answer: \(\displaystyle \dfrac17\ln\left|\dfrac{x-2}{x+2}\right|+\dfrac{\sqrt3}{7}\tan^{-1}\dfrac{x}{\sqrt3}+C\)

  6. Exercise 36

    ∫x2dx(x2+a2)(x2+b2)\displaystyle \int \frac{x^2 d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}

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    NCERT’s answer
    \(\displaystyle \frac{1}{a^2-b^2}\left\{a \tan ^{-1} \frac{x}{a}-b \tan ^{-1} \frac{x}{b}\right\}+c\)
    Take \(\displaystyle a\neq b\). Split into partial fractions in \(\displaystyle x^2\). \[\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\frac{A}{x^2+a^2}+\frac{B}{x^2+b^2} \] \[x^2=A(x^2+b^2)+B(x^2+a^2) \] \[x^2=-a^2:\; -a^2=A(b^2-a^2) \Rightarrow A=\frac{a^2}{a^2-b^2} \] \[x^2=-b^2:\; -b^2=B(a^2-b^2) \Rightarrow B=\frac{-b^2}{a^2-b^2} \] \[I=\frac{a^2}{a^2-b^2}\int\frac{dx}{x^2+a^2}-\frac{b^2}{a^2-b^2}\int\frac{dx}{x^2+b^2} \] \[I=\frac{a^2}{a^2-b^2}\cdot\frac1a\tan^{-1}\frac xa-\frac{b^2}{a^2-b^2}\cdot\frac1b\tan^{-1}\frac xb+C \] Answer: \(\displaystyle \dfrac{1}{a^2-b^2}\left[a\tan^{-1}\dfrac xa-b\tan^{-1}\dfrac xb\right]+C\)

  7. Exercise 37

    ∫0πx1+sin⁡x\displaystyle \int_0^\pi \frac{x}{1+\sin x}

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    NCERT’s answer
    \(\displaystyle \pi\)
    Use \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\) with \(\displaystyle \sin(\pi-x)=\sin x\). \[I=\int_0^\pi\frac{x\,dx}{1+\sin x}=\int_0^\pi\frac{(\pi-x)\,dx}{1+\sin x} \] \[2I=\pi\int_0^\pi\frac{dx}{1+\sin x} \] Since \(\displaystyle 1+\sin x=1+\cos\left(\tfrac\pi2-x\right)=2\cos^2\left(\tfrac\pi4-\tfrac x2\right)\): \[2I=\frac\pi2\int_0^\pi\sec^2\left(\frac\pi4-\frac x2\right)dx=\frac\pi2\Big[-2\tan\left(\frac\pi4-\frac x2\right)\Big]_0^\pi \] \[2I=\pi\Big[-\tan\left(-\frac\pi4\right)+\tan\frac\pi4\Big]=2\pi \] \[I=\pi \] Answer: \(\displaystyle \pi\)

  8. Exercise 38

    ∫2x−1(x−1)(x+2)(x−3)dx\displaystyle \int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x

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    NCERT’s answer
    \(\displaystyle \log \left|\frac{\sqrt{x-3}}{(x-1)^{\frac{1}{6}}(x+2)^{\frac{1}{3}}}\right|+c\)
    Partial fractions. \[\frac{2x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3} \] \[x=1:\; A=\frac{1}{(3)(-2)}=-\frac16 \] \[x=-2:\; B=\frac{-5}{(-3)(-5)}=-\frac13 \] \[x=3:\; C=\frac{5}{(2)(5)}=\frac12 \] \[I=-\frac16\int\frac{dx}{x-1}-\frac13\int\frac{dx}{x+2}+\frac12\int\frac{dx}{x-3} \] Answer: \(\displaystyle -\dfrac16\ln|x-1|-\dfrac13\ln|x+2|+\dfrac12\ln|x-3|+C\)

  9. Exercise 39

    ∫etan⁡−1x(1+x+x21+x2)dx\displaystyle \int e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right) d x

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    NCERT’s answer
    \(\displaystyle x e^{\tan ^{-1} x}+c\)
    Put \(\displaystyle t=\tan^{-1}x\). \[x=\tan t,\quad dx=\sec^2t\,dt,\quad 1+x^2=\sec^2t \] \[I=\int e^{t}\,\frac{1+\tan t+\tan^2t}{\sec^2t}\,\sec^2t\,dt=\int e^{t}\left(\sec^2t+\tan t\right)dt \] Use \(\displaystyle \int e^t\left(f(t)+f'(t)\right)dt=e^tf(t)+C\) with \(\displaystyle f(t)=\tan t\), \(\displaystyle f'(t)=\sec^2t\): \[I=e^{t}\tan t+C \] Answer: \(\displaystyle x\,e^{\tan^{-1}x}+C\)

  10. Exercise 40

    ∫sin⁡−1xa+xdx\displaystyle \int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x (Hint: Put x=atan⁡2θ\displaystyle x=a \tan ^2 \theta)

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    NCERT’s answer
    \(\displaystyle a\left[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}\right]+c\)
    Put \(\displaystyle x=a\tan^2\theta\), \(\displaystyle 0\le\theta<\tfrac\pi2\). \[\frac{x}{a+x}=\frac{\tan^2\theta}{1+\tan^2\theta}=\sin^2\theta \;\Rightarrow\; \sin^{-1}\sqrt{\frac{x}{a+x}}=\theta \] \[dx=2a\tan\theta\sec^2\theta\,d\theta \] \[I=\int\theta\cdot 2a\tan\theta\sec^2\theta\,d\theta \] By parts, with \(\displaystyle \int 2a\tan\theta\sec^2\theta\,d\theta=a\tan^2\theta\): \[I=a\theta\tan^2\theta-a\int\tan^2\theta\,d\theta \] \[I=a\theta\tan^2\theta-a(\tan\theta-\theta)+C=a\theta\sec^2\theta-a\tan\theta+C \] Back-substitute \(\displaystyle \tan\theta=\sqrt{x/a}\), \(\displaystyle \sec^2\theta=\dfrac{a+x}{a}\): \[I=(a+x)\tan^{-1}\sqrt{\frac xa}-\sqrt{ax}+C \] Answer: \(\displaystyle (a+x)\tan^{-1}\sqrt{\dfrac xa}-\sqrt{ax}+C\)