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NCERT Exemplar · Class 12 Mathematics Integrals

63 questions · 63 still being checked

EXERCISE 7.3 51–63 (part 6 of 6)

  1. Choose the correct option from given four options in each of the Exercises from $\displaystyle 48$ to 63.

    Exercise 51

    ∫ex(1−x1+x2)2dx\displaystyle \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x is equal to
    (A)
    ex1+x2+C\displaystyle \frac{e^x}{1+x^2}+\mathrm{C}
    (B)
    −ex1+x2+C\displaystyle \frac{-e^x}{1+x^2}+\mathrm{C}
    (C)
    ex(1+x2)2+C\displaystyle \frac{e^x}{\left(1+x^2\right)^2}+\mathrm{C}
    (D)
    −ex(1+x2)2+C\displaystyle \frac{-e^x}{\left(1+x^2\right)^2}+\mathrm{C}

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    (A) \(\displaystyle \dfrac{e^x}{1+x^2}+\mathrm{C}\)Split the fraction into the form \(\displaystyle f(x)+f'(x)\).\[\left(\frac{1-x}{1+x^2}\right)^2=\frac{(1+x^2)-2x}{(1+x^2)^2}=\frac{1}{1+x^2}-\frac{2x}{(1+x^2)^2} \]\[f(x)=\frac{1}{1+x^2}\ \Rightarrow\ f'(x)=-\frac{2x}{(1+x^2)^2} \]\[\int e^x\big[f(x)+f'(x)\big]dx=e^xf(x)+\mathrm{C}=\frac{e^x}{1+x^2}+\mathrm{C} \]

  2. Exercise 52

    ∫x9(4x2+1)6dx\displaystyle \int \frac{x^9}{\left(4 x^2+1\right)^6} d x is equal to
    (A)
    15x(4+1x2)−5+C\displaystyle \frac{1}{5 x}\left(4+\frac{1}{x^2}\right)^{-5}+\mathrm{C}
    (B)
    15(4+1x2)−5+C\displaystyle \frac{1}{5}\left(4+\frac{1}{x^2}\right)^{-5}+\mathrm{C}
    (C)
    110x(1+4)−5+C\displaystyle \frac{1}{10 x}(1+4)^{-5}+\mathrm{C}
    (D)
    110(1x2+4)−5+C\displaystyle \frac{1}{10}\left(\frac{1}{x^2}+4\right)^{-5}+\mathrm{C}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{1}{10}\left(\dfrac{1}{x^2}+4\right)^{-5}+\mathrm{C}\)Take \(\displaystyle x^{12}\) out of the denominator, then substitute.\[\frac{x^9}{(4x^2+1)^6}=\frac{x^9}{x^{12}\left(4+\dfrac{1}{x^2}\right)^6}=\frac{1}{x^3}\left(4+\frac{1}{x^2}\right)^{-6} \]\[t=4+\frac{1}{x^2}\ \Rightarrow\ dt=-\frac{2}{x^3}\,dx \]\[\int\frac{1}{x^3}\left(4+\frac{1}{x^2}\right)^{-6}dx=-\frac12\int t^{-6}\,dt \]\[=\frac{1}{10}\,t^{-5}+\mathrm{C}=\frac{1}{10}\left(\frac{1}{x^2}+4\right)^{-5}+\mathrm{C} \]

  3. Exercise 53

    If ∫dx(x+2)(x2+1)=alog⁡∣1+x2∣+btan⁡−1x+15log⁡∣x+2∣+C\displaystyle \int \frac{d x}{(x+2)\left(x^2+1\right)}=a \log \left|1+x^2\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+\mathrm{C}, then
    (A)
    a=−110,b=−25\displaystyle a=\frac{-1}{10}, b=\frac{-2}{5}
    (B)
    a=110,b=−25\displaystyle a=\frac{1}{10}, b=-\frac{2}{5}
    (C)
    a=−110,b=25\displaystyle a=\frac{-1}{10}, b=\frac{2}{5}
    (D)
    a=110,b=25\displaystyle a=\frac{1}{10}, b=\frac{2}{5}

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    NCERT’s answer
    C
    (C) \(\displaystyle a=-\dfrac{1}{10},\ b=\dfrac{2}{5}\)Resolve into partial fractions and integrate.\[\frac{1}{(x+2)(x^2+1)}=\frac{A}{x+2}+\frac{Bx+D}{x^2+1} \]\[1=A(x^2+1)+(Bx+D)(x+2) \]\[x=-2:\quad 1=5A\ \Rightarrow\ A=\frac15 \]\[x^2:\quad A+B=0\ \Rightarrow\ B=-\frac15 \]\[x^0:\quad A+2D=1\ \Rightarrow\ D=\frac25 \]\[\int=\frac15\log|x+2|-\frac{1}{10}\int\frac{2x\,dx}{x^2+1}+\frac25\int\frac{dx}{x^2+1} \]\[=\frac15\log|x+2|-\frac{1}{10}\log(1+x^2)+\frac25\tan^{-1}x+\mathrm{C} \]

  4. Exercise 54

    ∫x3x+1\displaystyle \int \frac{x^3}{x+1} is equal to
    (A)
    x+x22+x33−log⁡∣1−x∣+C\displaystyle x+\frac{x^2}{2}+\frac{x^3}{3}-\log |1-x|+\mathrm{C}
    (B)
    x+x22−x33−log⁡∣1−x∣+C\displaystyle x+\frac{x^2}{2}-\frac{x^3}{3}-\log |1-x|+\mathrm{C}
    (C)
    x−x22−x33−log⁡∣1+x∣+C\displaystyle x-\frac{x^2}{2}-\frac{x^3}{3}-\log |1+x|+\mathrm{C}
    (D)
    x−x22+x33−log⁡∣1+x∣+C\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3}-\log |1+x|+\mathrm{C}

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    NCERT’s answer
    D
    (D) \(\displaystyle x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\log|1+x|+\mathrm{C}\)Divide \(\displaystyle x^3\) by \(\displaystyle x+1\).\[x^3=(x+1)(x^2-x+1)-1 \]\[\frac{x^3}{x+1}=x^2-x+1-\frac{1}{x+1} \]\[\int\frac{x^3}{x+1}\,dx=\frac{x^3}{3}-\frac{x^2}{2}+x-\log|x+1|+\mathrm{C} \]

  5. Exercise 55

    ∫x+sin⁡x1+cos⁡xdx\displaystyle \int \frac{x+\sin x}{1+\cos x} d x is equal to
    (A)
    log⁡∣1+cos⁡x∣+C\displaystyle \log |1+\cos x|+\mathrm{C}
    (B)
    log⁡∣x+sin⁡x∣+C\displaystyle \log |x+\sin x|+\mathrm{C}
    (C)
    x−tan⁡x2+C\displaystyle x-\tan \frac{x}{2}+\mathrm{C}
    (D)
    x.tan⁡x2+C\displaystyle x . \tan \frac{x}{2}+\mathrm{C}

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    NCERT’s answer
    D
    (D) \(\displaystyle x\tan\dfrac{x}{2}+\mathrm{C}\)Use half-angle forms, then integrate the first term by parts.\[1+\cos x=2\cos^2\frac x2,\qquad \sin x=2\sin\frac x2\cos\frac x2 \]\[\frac{x+\sin x}{1+\cos x}=\frac{x}{2}\sec^2\frac x2+\tan\frac x2 \]\[\int\frac x2\sec^2\frac x2\,dx=x\tan\frac x2-\int\tan\frac x2\,dx \]\[\int\frac{x+\sin x}{1+\cos x}\,dx=x\tan\frac x2-\int\tan\frac x2\,dx+\int\tan\frac x2\,dx=x\tan\frac x2+\mathrm{C} \]

  6. Exercise 56

    If ∫x3dx1+x2=a(1+x2)32+b1+x2+C\displaystyle \int \frac{x^3 d x}{\sqrt{1+x^2}}=a\left(1+x^2\right)^{\frac{3}{2}}+b \sqrt{1+x^2}+\mathrm{C}, then
    (A)
    a=13,b=1\displaystyle a=\frac{1}{3}, \quad b=1
    (B)
    a=−13,b=1\displaystyle a=\frac{-1}{3}, \quad b=1
    (C)
    a=−13,b=−1\displaystyle a=\frac{-1}{3}, \quad b=-1
    (D)
    a=13,b=−1\displaystyle a=\frac{1}{3}, \quad b=-1

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    NCERT’s answer
    D
    (D) \(\displaystyle a=\dfrac13,\ b=-1\)Substitute \(\displaystyle t=1+x^2\).\[t=1+x^2,\qquad x\,dx=\frac{dt}{2},\qquad x^2=t-1 \]\[\int\frac{x^3\,dx}{\sqrt{1+x^2}}=\frac12\int\frac{t-1}{\sqrt t}\,dt=\frac12\int\left(t^{1/2}-t^{-1/2}\right)dt \]\[=\frac13\,t^{3/2}-t^{1/2}+\mathrm{C} \]\[=\frac13(1+x^2)^{3/2}-\sqrt{1+x^2}+\mathrm{C} \]

  7. Exercise 57

    ∫−π4π4dx1+cos⁡2x\displaystyle \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{d} x}{1+\cos 2 x} is equal to
    (A)
    1\displaystyle 1 (B) 2\displaystyle 2 (C) 3\displaystyle 3 (D) 4\displaystyle 4

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    NCERT’s answer
    A
    (A) \(\displaystyle 1\), because \(\displaystyle 1+\cos 2x = 2\cos^2 x\).\[\int_{-\pi/4}^{\pi/4} \frac{dx}{1+\cos 2x} = \int_{-\pi/4}^{\pi/4} \frac{dx}{2\cos^2 x} \] \[= \frac12\int_{-\pi/4}^{\pi/4} \sec^2 x\,dx = \frac12\left[\tan x\right]_{-\pi/4}^{\pi/4} \] \[= \frac12\big(1-(-1)\big) = 1 \]

  8. Exercise 58

    ∫0π21−sin⁡2xdx\displaystyle \int_0^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} d x is equal to
    (A)
    22\displaystyle 2 \sqrt{2}
    (B)
    2(2+1)\displaystyle 2(\sqrt{2}+1)
    (C)
    2\displaystyle 2 (D) 2(2−1)\displaystyle 2(\sqrt{2}-1)

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    NCERT’s answer
    D
    (D) \(\displaystyle 2(\sqrt2-1)\), because the integrand is \(\displaystyle |\cos x-\sin x|\), which changes sign at \(\displaystyle x=\frac{\pi}{4}\).\[1-\sin 2x = (\cos x-\sin x)^2 \] \[\sqrt{1-\sin 2x} = |\cos x-\sin x| \] \[I = \int_0^{\pi/4}(\cos x-\sin x)\,dx + \int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx \] \[= \left[\sin x+\cos x\right]_0^{\pi/4} + \left[-\cos x-\sin x\right]_{\pi/4}^{\pi/2} \] \[= (\sqrt2-1)+(\sqrt2-1) = 2(\sqrt2-1) \]

  9. Exercise 59

    ∫0π2cos⁡xesin⁡xdx\displaystyle \int_0^{\frac{\pi}{2}} \cos x e^{\sin x} d x is equal to ____.

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    NCERT’s answer
    \(\displaystyle e-1\)
    \(\displaystyle e-1\)\[t=\sin x,\qquad dt=\cos x\,dx \] \[x=0\Rightarrow t=0,\qquad x=\frac{\pi}{2}\Rightarrow t=1 \] \[\int_0^{\pi/2}\cos x\,e^{\sin x}\,dx = \int_0^1 e^t\,dt = \left[e^t\right]_0^1 = e-1 \]
  10. Fill in the blanks in each of the following Exercise $\displaystyle 60$ to 63.

    Exercise 60

    ∫x+3(x+4)2exdx=\displaystyle \int \frac{x+3}{(x+4)^2} e^x d x= ____.

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    NCERT’s answer
    \(\displaystyle \frac{e^x}{x+4}+c\)
    \(\displaystyle \dfrac{e^x}{x+4}+C\)\[\frac{x+3}{(x+4)^2} = \frac{(x+4)-1}{(x+4)^2} = \frac{1}{x+4}-\frac{1}{(x+4)^2} \] \[f(x)=\frac{1}{x+4},\qquad f'(x)=-\frac{1}{(x+4)^2} \] \[\int e^x\big(f(x)+f'(x)\big)\,dx = e^x f(x)+C \] \[\int \frac{x+3}{(x+4)^2}\,e^x\,dx = \frac{e^x}{x+4}+C \]

  11. Exercise 61

    If ∫0a11+4x2dx=π8\displaystyle \int_0^a \frac{1}{1+4 x^2} d x=\frac{\pi}{8}, then a=\displaystyle a= ____.

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\)
    \(\displaystyle a=\dfrac12\)\[\int_0^a \frac{dx}{1+4x^2} = \frac12\left[\tan^{-1}2x\right]_0^a = \frac12\tan^{-1}2a \] \[\frac12\tan^{-1}2a = \frac{\pi}{8} \] \[\tan^{-1}2a = \frac{\pi}{4} \] \[2a = \tan\frac{\pi}{4} = 1 \] \[a = \frac12 \]

  12. Exercise 62

    ∫sin⁡x3+4cos⁡2xdx=\displaystyle \int \frac{\sin x}{3+4 \cos ^2 x} d x= ____.

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    NCERT’s answer
    \(\displaystyle \frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c\)
    \(\displaystyle -\dfrac{1}{2\sqrt3}\tan^{-1}\!\left(\dfrac{2\cos x}{\sqrt3}\right)+C\)\[t=\cos x,\qquad dt=-\sin x\,dx \] \[I = -\int\frac{dt}{3+4t^2} = -\frac14\int\frac{dt}{t^2+\left(\dfrac{\sqrt3}{2}\right)^2} \] \[= -\frac14\cdot\frac{2}{\sqrt3}\,\tan^{-1}\frac{2t}{\sqrt3}+C \] \[= -\frac{1}{2\sqrt3}\tan^{-1}\!\left(\frac{2\cos x}{\sqrt3}\right)+C \]

  13. Exercise 63

    The value of ∫−ππsin⁡3xcos⁡2xdx\displaystyle \int_{-\pi}^\pi \sin ^3 x \cos ^2 x d x is ____.

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    NCERT’s answer
    $\displaystyle 0$
    \(\displaystyle 0\)\[f(x)=\sin^3x\cos^2x \] \[f(-x)=(-\sin x)^3\cos^2x=-f(x) \] \[\int_{-a}^{a}f(x)\,dx=0 \quad (f\text{ odd}) \] \[\int_{-\pi}^{\pi}\sin^3x\cos^2x\,dx=0 \]