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NCERT Solutions · Class 12 Mathematics Integrals

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Miscellaneous Exercise 31–40 (part 27 of 27)

  1. Evaluate the definite integrals in Exercises $\displaystyle 24$ to 31.

    Exercise 31

    ∫14[∣x−1∣+∣x−2∣+∣x−3∣]dx\displaystyle \int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x

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    NCERT’s answer
    \(\displaystyle \frac{19}{2}\)
    A modulus must be split at the point where its argument changes sign, and \(\displaystyle x=2\) and \(\displaystyle x=3\) both lie inside \(\displaystyle [1,4]\). On \(\displaystyle [1,4]\) the first modulus needs no split, since \(\displaystyle x-1\ge 0\) throughout. \[\int_{1}^{4}|x-1|\,dx=\int_{1}^{4}(x-1)dx=\left[\frac{(x-1)^{2}}{2}\right]_{1}^{4}=\frac92. \] \[\int_{1}^{4}|x-2|\,dx=\int_{1}^{2}(2-x)dx+\int_{2}^{4}(x-2)dx=\frac12+2=\frac52. \] \[\int_{1}^{4}|x-3|\,dx=\int_{1}^{3}(3-x)dx+\int_{3}^{4}(x-3)dx=2+\frac12=\frac52. \] Adding the three, \[\int_{1}^{4}\big[|x-1|+|x-2|+|x-3|\big]dx=\frac92+\frac52+\frac52=\frac{19}{2}. \]
  2. Prove the following (Exercises $\displaystyle 32$ to $\displaystyle 37$)

    Exercise 32

    ∫13dxx2(x+1)=23+log⁡23\displaystyle \int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}

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    The denominator carries the repeated factor \(\displaystyle x^{2}\), so the partial-fraction form is \[\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1},\qquad 1=Ax(x+1)+B(x+1)+Cx^{2}. \] Putting \(\displaystyle x=0\): \(\displaystyle B=1\). Putting \(\displaystyle x=-1\): \(\displaystyle C=1\). Comparing coefficients of \(\displaystyle x^{2}\): \(\displaystyle 0=A+C\), so \(\displaystyle A=-1\). Hence \[\int\frac{dx}{x^{2}(x+1)}=-\log|x|-\frac1x+\log|x+1|=\log\left|\frac{x+1}{x}\right|-\frac1x. \] Both limits are positive, so the moduli may be dropped: \[\int_{1}^{3}\frac{dx}{x^{2}(x+1)}=\left(\log\frac43-\frac13\right)-\left(\log 2-1\right)=\log\frac{4}{3\cdot 2}+\left(1-\frac13\right)=\frac23+\log\frac23. \] This is the required result, \(\displaystyle \displaystyle\int_{1}^{3}\frac{dx}{x^{2}(x+1)}=\frac23+\log\frac23\).

  3. Exercise 33

    ∫01xexdx=1\displaystyle \int_{0}^{1} x e^{x} d x=1

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    Integrate by parts, taking \(\displaystyle x\) as the first function and \(\displaystyle e^{x}\) as the second (ILATE: algebraic before exponential): \[\int_{0}^{1}xe^{x}dx=\left[xe^{x}\right]_{0}^{1}-\int_{0}^{1}1\cdot e^{x}dx=\left[xe^{x}-e^{x}\right]_{0}^{1}. \] Evaluating, \[=\left(1\cdot e-e\right)-\left(0\cdot 1-1\right)=0-(-1)=1. \] Hence \(\displaystyle \displaystyle\int_{0}^{1}xe^{x}dx=1\), as required.

  4. Exercise 34

    ∫−11x17cos⁡4xdx=0\displaystyle \int_{-1}^{1} x^{17} \cos ^{4} x d x=0

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    Use the property that \(\displaystyle \int_{-a}^{a}f(x)dx=0\) whenever \(\displaystyle f\) is odd, i.e. \(\displaystyle f(-x)=-f(x)\). Let \(\displaystyle f(x)=x^{17}\cos^{4}x\). Then, since \(\displaystyle 17\) is odd and \(\displaystyle \cos\) is an even function, \[f(-x)=(-x)^{17}\cos^{4}(-x)=-x^{17}\cos^{4}x=-f(x). \] So \(\displaystyle f\) is odd, and the interval \(\displaystyle [-1,1]\) is symmetric about the origin. Therefore \[\int_{-1}^{1}x^{17}\cos^{4}x\,dx=0. \]

  5. Exercise 35

    ∫0π2sin⁡3xdx=23\displaystyle \int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}

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    Split off one factor of \(\displaystyle \sin x\) and convert the rest, using \(\displaystyle \sin^{2}x=1-\cos^{2}x\): \[\sin^{3}x=\sin x\left(1-\cos^{2}x\right). \] Put \(\displaystyle t=\cos x\), so \(\displaystyle dt=-\sin x\,dx\); the limits become \(\displaystyle x=0\Rightarrow t=1\) and \(\displaystyle x=\frac{\pi}{2}\Rightarrow t=0\) (note the reversal, which the minus sign cancels): \[\int_{0}^{\pi/2}\sin^{3}x\,dx=-\int_{1}^{0}\left(1-t^{2}\right)dt=\int_{0}^{1}\left(1-t^{2}\right)dt=\left[t-\frac{t^{3}}{3}\right]_{0}^{1}=1-\frac13=\frac23. \] Hence \(\displaystyle \displaystyle\int_{0}^{\pi/2}\sin^{3}x\,dx=\frac23\), as required.

  6. Exercise 36

    ∫0π42tan⁡3xdx=1−log⁡2\displaystyle \int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2

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    Reduce the power using \(\displaystyle \tan^{2}x=\sec^{2}x-1\): \[\tan^{3}x=\tan x\left(\sec^{2}x-1\right)=\tan x\sec^{2}x-\tan x. \] With \(\displaystyle \int\tan x\sec^{2}x\,dx=\frac{\tan^{2}x}{2}\) (put \(\displaystyle u=\tan x\)) and \(\displaystyle \int\tan x\,dx=-\log|\cos x|\), \[\int 2\tan^{3}x\,dx=2\left[\frac{\tan^{2}x}{2}+\log|\cos x|\right]=\tan^{2}x+2\log|\cos x|. \] Evaluating between \(\displaystyle 0\) and \(\displaystyle \frac{\pi}{4}\), where \(\displaystyle \tan\frac{\pi}{4}=1\) and \(\displaystyle \cos\frac{\pi}{4}=\frac{1}{\sqrt2}\), \[\int_{0}^{\pi/4}2\tan^{3}x\,dx=\left(1+2\log\frac{1}{\sqrt2}\right)-\left(0+2\log 1\right)=1-\log 2. \] Hence \(\displaystyle \displaystyle\int_{0}^{\pi/4}2\tan^{3}x\,dx=1-\log 2\), as required.

  7. Exercise 37

    ∫01sin⁡−1xdx=π2−1\displaystyle \int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1

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    Integrate by parts, taking \(\displaystyle \sin^{-1}x\) as the first function and \(\displaystyle 1\) as the second: \[\int_{0}^{1}\sin^{-1}x\,dx=\left[x\sin^{-1}x\right]_{0}^{1}-\int_{0}^{1}\frac{x}{\sqrt{1-x^{2}}}dx. \] The first bracket is \(\displaystyle 1\cdot\frac{\pi}{2}-0=\frac{\pi}{2}\). For the remaining integral put \(\displaystyle u=1-x^{2}\), \(\displaystyle du=-2x\,dx\), giving \[\int_{0}^{1}\frac{x\,dx}{\sqrt{1-x^{2}}}=\left[-\sqrt{1-x^{2}}\right]_{0}^{1}=0-(-1)=1. \] Hence \[\int_{0}^{1}\sin^{-1}x\,dx=\frac{\pi}{2}-1, \] as required.
  8. Choose the correct answers in Exercises $\displaystyle 38$ to $\displaystyle 40$

    Exercise 38

    ∫dxex+e−x\displaystyle \int \frac{d x}{e^{x}+e^{-x}} is equal to (A) tan⁡−1(ex)+C\displaystyle \tan ^{-1}\left(e^{x}\right)+\mathrm{C} (B) tan⁡−1(e−x)+C\displaystyle \tan ^{-1}\left(e^{-x}\right)+\mathrm{C} (C) log⁡(ex−e−x)+C\displaystyle \log \left(e^{x}-e^{-x}\right)+\mathrm{C} (D) log⁡(ex+e−x)+C\displaystyle \log \left(e^{x}+e^{-x}\right)+\mathrm{C}

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    NCERT’s answer
    A
    Multiply numerator and denominator by \(\displaystyle e^{x}\), so that the derivative needed for the substitution appears: \[\int\frac{dx}{e^{x}+e^{-x}}=\int\frac{e^{x}\,dx}{e^{2x}+1}. \] Put \(\displaystyle t=e^{x}\), so \(\displaystyle dt=e^{x}dx\) and \(\displaystyle e^{2x}=t^{2}\): \[=\int\frac{dt}{1+t^{2}}=\tan^{-1}t+\mathrm{C}=\tan^{-1}\left(e^{x}\right)+\mathrm{C}. \] So the correct answer is (A).

  9. Exercise 39

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

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    NCERT’s answer
    B
    Factorise the numerator, \(\displaystyle \cos 2x=\cos^{2}x-\sin^{2}x=(\cos x-\sin x)(\cos x+\sin x)\), and cancel one factor of \(\displaystyle (\sin x+\cos x)\): \[\int\frac{\cos 2x}{(\sin x+\cos x)^{2}}dx=\int\frac{\cos x-\sin x}{\sin x+\cos x}dx. \] The numerator is now exactly the derivative of the denominator, so put \(\displaystyle t=\sin x+\cos x\), \(\displaystyle dt=(\cos x-\sin x)dx\): \[=\int\frac{dt}{t}=\log|t|+\mathrm{C}=\log|\sin x+\cos x|+\mathrm{C}. \] So the correct answer is (B).

  10. Exercise 40

    If f(a+b−x)=f(x)\displaystyle f(a+b-x)=f(x), then ∫abxf(x)dx\displaystyle \int_{a}^{b} x f(x) d x is equal to (A) a+b2∫abf(b−x)dx\displaystyle \frac{a+b}{2} \int_{a}^{b} f(b-x) d x (B) a+b2∫abf(b+x)dx\displaystyle \frac{a+b}{2} \int_{a}^{b} f(b+x) d x (C) b−a2∫abf(x)dx\displaystyle \frac{b-a}{2} \int_{a}^{b} f(x) d x (D) a+b2∫abf(x)dx\displaystyle \frac{a+b}{2} \int_{a}^{b} f(x) d x

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    NCERT’s answer
    D
    Use the property \(\displaystyle \displaystyle\int_{a}^{b}g(x)\,dx=\int_{a}^{b}g(a+b-x)\,dx\). Writing \(\displaystyle I\) for the required integral, \[I=\int_{a}^{b}xf(x)\,dx=\int_{a}^{b}(a+b-x)\,f(a+b-x)\,dx. \] The hypothesis \(\displaystyle f(a+b-x)=f(x)\) now replaces the second factor: \[I=\int_{a}^{b}(a+b-x)f(x)\,dx=(a+b)\int_{a}^{b}f(x)\,dx-\int_{a}^{b}xf(x)\,dx=(a+b)\int_{a}^{b}f(x)\,dx-I. \] Bringing \(\displaystyle I\) to the left, \(\displaystyle 2I=(a+b)\displaystyle\int_{a}^{b}f(x)\,dx\), so \[\int_{a}^{b}xf(x)\,dx=\frac{a+b}{2}\int_{a}^{b}f(x)\,dx. \] So the correct answer is (D).