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NCERT Exemplar · Class 12 Mathematics Integrals

63 questions · 63 still being checked

EXERCISE 7.3 41–50 (part 5 of 6)

  1. Evaluate the following:

    Exercise 41

    ∫π3π21+cos⁡x(1−cos⁡x)52\displaystyle \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}}

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    NCERT’s answer
    \(\displaystyle \frac{3}{2}\)
    Half-angle forms: \[1+\cos x = 2\cos^2\tfrac x2, \qquad 1-\cos x = 2\sin^2\tfrac x2 \] \[I = \int_{\pi/3}^{\pi/2} \frac{\sqrt2\,\cos\frac x2}{4\sqrt2\,\sin^5\frac x2}\,dx = \frac14\int_{\pi/3}^{\pi/2} \frac{\cos\frac x2}{\sin^5\frac x2}\,dx \] Put \(\displaystyle t=\sin\frac x2\): \[\cos\tfrac x2\,dx = 2\,dt, \qquad x=\tfrac\pi3 \to t=\tfrac12, \qquad x=\tfrac\pi2 \to t=\tfrac1{\sqrt2} \] \[I = \frac12\int_{1/2}^{1/\sqrt2} t^{-5}\,dt = \left[-\frac{1}{8t^{4}}\right]_{1/2}^{1/\sqrt2} \] \[I = -\frac48 + \frac{16}{8} = \frac32 \] Answer: \(\displaystyle \dfrac32\)

  2. Exercise 42

    ∫e−3xcos⁡3xdx\displaystyle \int e^{-3 x} \cos ^3 x d x

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    NCERT’s answer
    \(\displaystyle \frac{e^{-3 x}}{24}[\sin 3 x-\cos 3 x]+\frac{3 e^{-3 x}}{40}[\sin x-3 \cos x]+c\)
    Reduce the power: \[\cos^3 x = \tfrac14(\cos 3x + 3\cos x) \] \[\int e^{ax}\cos bx\,dx = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2+b^2}+C \] With \(\displaystyle a=-3\): \[\int e^{-3x}\cos 3x\,dx = \frac{e^{-3x}(\sin 3x-\cos 3x)}{6} \] \[\int e^{-3x}\cos x\,dx = \frac{e^{-3x}(\sin x-3\cos x)}{10} \] \[I = \frac14\left[\frac{e^{-3x}(\sin 3x-\cos 3x)}{6}+\frac{3e^{-3x}(\sin x-3\cos x)}{10}\right]+C \] Answer: \(\displaystyle \dfrac{e^{-3x}}{120}\left(5\sin 3x-5\cos 3x+9\sin x-27\cos x\right)+C\)

  3. Exercise 43

    ∫tan⁡xdx\displaystyle \int \sqrt{\tan x} d x (Hint: Put tan⁡x=t2\displaystyle \tan x=t^2)

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    NCERT’s answer
    \(\displaystyle \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+\frac{1}{2 \sqrt{2}} \log \left|\frac{\tan x-\sqrt{2 \tan x}+1}{\tan x+\sqrt{2 \tan x}+1}\right|+c\)
    Put \(\displaystyle \tan x=t^2\): \[\sec^2x\,dx = 2t\,dt \;\Rightarrow\; dx=\frac{2t\,dt}{1+t^4} \] \[I = \int t\cdot\frac{2t}{1+t^4}\,dt = \int \frac{2t^2}{1+t^4}\,dt, \qquad 2t^2=(t^2+1)+(t^2-1) \] Divide numerator and denominator by \(\displaystyle t^2\): \[I = \int \frac{1+\frac1{t^2}}{t^2+\frac1{t^2}}\,dt + \int \frac{1-\frac1{t^2}}{t^2+\frac1{t^2}}\,dt \] With \(\displaystyle u=t-\frac1t\) and \(\displaystyle v=t+\frac1t\): \[du=\left(1+\tfrac1{t^2}\right)dt,\quad dv=\left(1-\tfrac1{t^2}\right)dt,\quad t^2+\tfrac1{t^2}=u^2+2=v^2-2 \] \[I = \int\frac{du}{u^2+2}+\int\frac{dv}{v^2-2} = \frac1{\sqrt2}\tan^{-1}\frac{u}{\sqrt2}+\frac1{2\sqrt2}\log\left|\frac{v-\sqrt2}{v+\sqrt2}\right|+C \] \[\frac{u}{\sqrt2}=\frac{t^2-1}{\sqrt2\,t},\qquad \frac{v-\sqrt2}{v+\sqrt2}=\frac{t^2-\sqrt2\,t+1}{t^2+\sqrt2\,t+1} \] Answer: \(\displaystyle \dfrac1{\sqrt2}\tan^{-1}\dfrac{\tan x-1}{\sqrt{2\tan x}}+\dfrac1{2\sqrt2}\log\dfrac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}+C\)

  4. Exercise 44

    ∫0π2dx(a2cos⁡2x+b2sin⁡2x)2\displaystyle \int_0^{\frac{\pi}{2}} \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} (Hint: Divide Numerator and Denominator by cos⁡4x\displaystyle \cos ^4 x)

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{4}\left(\frac{a^2+b^2}{a^3 b^3}\right)\)
    Divide numerator and denominator by \(\displaystyle \cos^4x\) (taking \(\displaystyle a,b>0\)): \[I = \int_0^{\pi/2}\frac{\sec^2x\,(1+\tan^2x)}{(a^2+b^2\tan^2x)^2}\,dx = \int_0^{\infty}\frac{1+t^2}{(a^2+b^2t^2)^2}\,dt \qquad (t=\tan x) \] Put \(\displaystyle bt=a\tan\theta\); \(\displaystyle \theta\) runs from \(\displaystyle 0\) to \(\displaystyle \frac\pi2\): \[dt=\frac ab\sec^2\theta\,d\theta,\qquad a^2+b^2t^2=a^2\sec^2\theta,\qquad 1+t^2=\frac{b^2+a^2\tan^2\theta}{b^2} \] \[I = \frac1{a^3b^3}\int_0^{\pi/2}\left(b^2\cos^2\theta+a^2\sin^2\theta\right)d\theta \] \[\int_0^{\pi/2}\cos^2\theta\,d\theta=\int_0^{\pi/2}\sin^2\theta\,d\theta=\frac\pi4 \] \[I = \frac{1}{a^3b^3}\cdot\frac\pi4\,(a^2+b^2) \] Answer: \(\displaystyle \dfrac{\pi(a^2+b^2)}{4a^3b^3}\)

  5. Exercise 45

    ∫01xlog⁡(1+2x)dx\displaystyle \int_0^1 x \log (1+2 x) d x

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    NCERT’s answer
    \(\displaystyle \frac{3}{8} \log 3\)
    Integrate by parts with \(\displaystyle u=\log(1+2x)\), \(\displaystyle dv=x\,dx\): \[I = \left[\frac{x^2}{2}\log(1+2x)\right]_0^1-\int_0^1\frac{x^2}{2}\cdot\frac{2}{1+2x}\,dx = \frac12\log3-\int_0^1\frac{x^2}{1+2x}\,dx \] \[\frac{x^2}{1+2x}=\frac x2-\frac14+\frac14\cdot\frac1{1+2x} \] \[\int_0^1\frac{x^2}{1+2x}\,dx=\left[\frac{x^2}{4}-\frac x4+\frac18\log(1+2x)\right]_0^1=\frac18\log3 \] \[I=\frac12\log3-\frac18\log3=\frac38\log3 \] Answer: \(\displaystyle \dfrac38\log3\)

  6. Exercise 46

    ∫0πxlog⁡sin⁡xdx\displaystyle \int_0^\pi x \log \sin x d x

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    NCERT’s answer
    \(\displaystyle \frac{\pi^2}{2} \log \frac{1}{2}\)
    Replace \(\displaystyle x\) by \(\displaystyle \pi-x\), using \(\displaystyle \sin(\pi-x)=\sin x\): \[I=\int_0^\pi x\log\sin x\,dx=\int_0^\pi(\pi-x)\log\sin x\,dx \] Add the two forms: \[2I=\pi\int_0^\pi\log\sin x\,dx=2\pi\int_0^{\pi/2}\log\sin x\,dx=2\pi J \] For \(\displaystyle J\), replace \(\displaystyle x\) by \(\displaystyle \frac\pi2-x\): \[J=\int_0^{\pi/2}\log\sin x\,dx=\int_0^{\pi/2}\log\cos x\,dx \] \[2J=\int_0^{\pi/2}\log\left(\tfrac12\sin2x\right)dx=\int_0^{\pi/2}\log\sin 2x\,dx-\frac\pi2\log2 \] \[\int_0^{\pi/2}\log\sin 2x\,dx=\frac12\int_0^{\pi}\log\sin w\,dw=J \;\Rightarrow\; J=-\frac\pi2\log2 \] \[I=\pi J=-\frac{\pi^2}{2}\log2 \] Answer: \(\displaystyle -\dfrac{\pi^2}{2}\log2\)

  7. Exercise 47

    ∫−π4π4log⁡(sin⁡x+cos⁡x)dx\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log (\sin x+\cos x) d x

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{4} \log \frac{1}{2}\)
    For an integral over \(\displaystyle [-a,a]\): \[\int_{-a}^{a}f(x)\,dx=\int_0^{a}\big[f(x)+f(-x)\big]dx \] \[f(x)+f(-x)=\log\big[(\cos x+\sin x)(\cos x-\sin x)\big]=\log\cos2x \] \[I=\int_0^{\pi/4}\log\cos2x\,dx=\frac12\int_0^{\pi/2}\log\cos u\,du=\frac12J \qquad (u=2x) \] Here \(\displaystyle J=\int_0^{\pi/2}\log\cos u\,du=\int_0^{\pi/2}\log\sin u\,du\), so \[2J=\int_0^{\pi/2}\log\left(\tfrac12\sin2u\right)du=\int_0^{\pi/2}\log\sin 2u\,du-\frac\pi2\log2 \] \[\int_0^{\pi/2}\log\sin 2u\,du=\frac12\int_0^{\pi}\log\sin w\,dw=J \;\Rightarrow\; J=-\frac\pi2\log2 \] \[I=\frac12J=-\frac\pi4\log2 \] Answer: \(\displaystyle -\dfrac\pi4\log2\)
  8. Choose the correct option from given four options in each of the Exercises from $\displaystyle 48$ to 63.

    Exercise 48

    ∫cos⁡2x−cos⁡2θcos⁡x−cos⁡θdx\displaystyle \int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x is equal to
    (A)
    2(sin⁡x+xcos⁡θ)+C\displaystyle 2(\sin x+x \cos \theta)+\mathrm{C}
    (B)
    2(sin⁡x−xcos⁡θ)+C\displaystyle 2(\sin x-x \cos \theta)+\mathrm{C}
    (C)
    2(sin⁡x+2xcos⁡θ)+C\displaystyle 2(\sin x+2 x \cos \theta)+\mathrm{C}
    (D)
    2(sin⁡x−2xcos⁡θ)+C\displaystyle 2(\sin x-2 x \cos \theta)+\mathrm{C}

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    NCERT’s answer
    A
    (A) \(\displaystyle 2(\sin x+x\cos\theta)+C\) \[\cos2x-\cos2\theta=2\cos^2x-2\cos^2\theta=2(\cos x-\cos\theta)(\cos x+\cos\theta) \] \[\int 2(\cos x+\cos\theta)\,dx=2(\sin x+x\cos\theta)+C \]

  9. Exercise 49

    ∫dxsin⁡(x−a)sin⁡(x−b)\displaystyle \int \frac{d x}{\sin (x-a) \sin (x-b)} is equal to
    (A)
    sin⁡(b−a)log⁡∣sin⁡(x−b)sin⁡(x−a)∣+C\displaystyle \sin (b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+\mathrm{C}
    (B)
    cosec⁡(b−a)log⁡∣sin⁡(x−a)sin⁡(x−b)∣+C\displaystyle \operatorname{cosec}(b-a) \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+\mathrm{C}
    (C)
    cosec⁡(b−a)log⁡∣sin⁡(x−b)sin⁡(x−a)∣+C\displaystyle \operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+\mathrm{C}
    (D)
    sin⁡(b−a)log⁡∣sin⁡(x−a)sin⁡(x−b)∣+C\displaystyle \sin (b-a) \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+\mathrm{C}

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    NCERT’s answer
    C
    (C) \(\displaystyle \operatorname{cosec}(b-a)\log\left|\dfrac{\sin(x-b)}{\sin(x-a)}\right|+\mathrm{C}\)Multiply and divide by \(\displaystyle \sin(b-a)\).\[\int\frac{dx}{\sin(x-a)\sin(x-b)}=\operatorname{cosec}(b-a)\int\frac{\sin(b-a)}{\sin(x-a)\sin(x-b)}\,dx \]\[\sin(b-a)=\sin\big((x-a)-(x-b)\big)=\sin(x-a)\cos(x-b)-\cos(x-a)\sin(x-b) \]\[\frac{\sin(b-a)}{\sin(x-a)\sin(x-b)}=\cot(x-b)-\cot(x-a) \]\[\int\frac{dx}{\sin(x-a)\sin(x-b)}=\operatorname{cosec}(b-a)\big[\log|\sin(x-b)|-\log|\sin(x-a)|\big]+\mathrm{C} \]\[=\operatorname{cosec}(b-a)\log\left|\frac{\sin(x-b)}{\sin(x-a)}\right|+\mathrm{C} \]

  10. Exercise 50

    ∫tan⁡−1xdx\displaystyle \int \tan ^{-1} \sqrt{x} d x is equal to
    (A)
    (x+1)tan⁡−1x−x+C\displaystyle (x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+\mathrm{C}
    (B)
    xtan⁡−1x−x+C\displaystyle x \tan ^{-1} \sqrt{x}-\sqrt{x}+\mathrm{C}
    (C)
    x−xtan⁡−1x+C\displaystyle \sqrt{x}-x \tan ^{-1} \sqrt{x}+\mathrm{C}
    (D)
    x−(x+1)tan⁡−1x+C\displaystyle \sqrt{x}-(x+1) \tan ^{-1} \sqrt{x}+\mathrm{C}

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    NCERT’s answer
    A
    (A) \(\displaystyle (x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+\mathrm{C}\)Substitute \(\displaystyle x=t^2\), then integrate by parts.\[x=t^2,\quad dx=2t\,dt \]\[\int\tan^{-1}\sqrt{x}\,dx=\int 2t\tan^{-1}t\,dt \]\[=t^2\tan^{-1}t-\int\frac{t^2}{1+t^2}\,dt \]\[\frac{t^2}{1+t^2}=1-\frac{1}{1+t^2} \]\[=t^2\tan^{-1}t-t+\tan^{-1}t+\mathrm{C} \]\[=(x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+\mathrm{C} \]