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NCERT Solutions · Class 12 Mathematics Integrals

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EXERCISE 7.10 1–10 (part 22 of 27)

  1. By using the properties of definite integrals, evaluate the integrals in Exercises $\displaystyle 1$ to 19.

    Exercise 1

    ∫0π2cos⁡2xdx\displaystyle \int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{4}\)
    Use the power-reduction identity \(\displaystyle \cos^{2}x=\frac{1+\cos 2x}{2}\), since \(\displaystyle \cos^{2}x\) has no elementary antiderivative in that form.\[\int_{0}^{\pi/2}\cos^{2}x\,dx=\frac{1}{2}\int_{0}^{\pi/2}\left(1+\cos 2x\right)dx=\frac{1}{2}\left[x+\frac{\sin 2x}{2}\right]_{0}^{\pi/2}.\]At \(\displaystyle x=\frac{\pi}{2}\) the bracket is \(\displaystyle \frac{\pi}{2}+\frac{\sin\pi}{2}=\frac{\pi}{2}\); at \(\displaystyle x=0\) it is \(\displaystyle 0\).\[=\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}.\]Hence \(\displaystyle \displaystyle\int_{0}^{\pi/2}\cos^{2}x\,dx=\frac{\pi}{4}\).

  2. Exercise 2

    ∫0π2sin⁡xsin⁡x+cos⁡xdx\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{4}\)
    Apply the property \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\frac{\pi}{2}\).Let \[I=\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.\]Replacing \(\displaystyle x\) by \(\displaystyle \frac{\pi}{2}-x\) interchanges sine and cosine, because \(\displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x\) and \(\displaystyle \cos\left(\frac{\pi}{2}-x\right)=\sin x\): \[I=\int_{0}^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.\]The two forms have the same denominator, so adding them adds only the numerators: \[2I=\int_{0}^{\pi/2}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx=\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}.\]Hence \(\displaystyle I=\dfrac{\pi}{4}\).

  3. Exercise 3

    ∫0π2sin⁡32xdxsin⁡32x+cos⁡32x\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{4}\)
    Same structure as the previous question: use \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\frac{\pi}{2}\).Let \[I=\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx.\]Under \(\displaystyle x\mapsto\frac{\pi}{2}-x\), \(\displaystyle \sin x\) and \(\displaystyle \cos x\) swap, so \[I=\int_{0}^{\pi/2}\frac{\cos^{3/2}x}{\cos^{3/2}x+\sin^{3/2}x}\,dx.\]The denominator is unchanged, so adding the two expressions gives \[2I=\int_{0}^{\pi/2}\frac{\sin^{3/2}x+\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx=\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}.\]Hence \(\displaystyle I=\dfrac{\pi}{4}\). (Notice the exponent never mattered — only that the two terms swap.)

  4. Exercise 4

    ∫0π2cos⁡5xdxsin⁡5x+cos⁡5x\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x d x}{\sin ^{5} x+\cos ^{5} x}

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{4}\)
    Use \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\frac{\pi}{2}\).Let \[I=\int_{0}^{\pi/2}\frac{\cos^{5}x}{\sin^{5}x+\cos^{5}x}\,dx.\]Replacing \(\displaystyle x\) by \(\displaystyle \frac{\pi}{2}-x\) swaps \(\displaystyle \sin x\) and \(\displaystyle \cos x\): \[I=\int_{0}^{\pi/2}\frac{\sin^{5}x}{\cos^{5}x+\sin^{5}x}\,dx.\]Adding the two (common denominator): \[2I=\int_{0}^{\pi/2}\frac{\cos^{5}x+\sin^{5}x}{\sin^{5}x+\cos^{5}x}\,dx=\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}.\]Hence \(\displaystyle I=\dfrac{\pi}{4}\).

  5. Exercise 5

    ∫−55∣x+2∣dx\displaystyle \int_{-5}^{5}|x+2| d x

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    NCERT’s answer
    $\displaystyle 29$
    The modulus changes its rule where \(\displaystyle x+2=0\), i.e. at \(\displaystyle x=-2\), and \(\displaystyle -2\) lies inside \(\displaystyle [-5,5]\). So split the integral there (additivity over adjacent intervals) — integrating \(\displaystyle |x+2|\) in one piece is the standard mistake here.On \(\displaystyle -5\le x\le-2\) we have \(\displaystyle x+2\le 0\), so \(\displaystyle |x+2|=-(x+2)\); on \(\displaystyle -2\le x\le 5\) we have \(\displaystyle x+2\ge 0\), so \(\displaystyle |x+2|=x+2\).\[\int_{-5}^{5}|x+2|\,dx=-\int_{-5}^{-2}(x+2)\,dx+\int_{-2}^{5}(x+2)\,dx.\]First piece: \[-\left[\frac{x^{2}}{2}+2x\right]_{-5}^{-2}=-\left[(2-4)-\left(\frac{25}{2}-10\right)\right]=-\left[-2-\frac{5}{2}\right]=\frac{9}{2}.\]Second piece: \[\left[\frac{x^{2}}{2}+2x\right]_{-2}^{5}=\left(\frac{25}{2}+10\right)-(2-4)=\frac{45}{2}+2=\frac{49}{2}.\]\[\int_{-5}^{5}|x+2|\,dx=\frac{9}{2}+\frac{49}{2}=29.\]

  6. Exercise 6

    ∫28∣x−5∣dx\displaystyle \int_{2}^{8}|x-5| d x

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    NCERT’s answer
    $\displaystyle 9$
    \(\displaystyle |x-5|\) changes rule at \(\displaystyle x=5\), which lies inside \(\displaystyle [2,8]\), so split the integral at \(\displaystyle 5\).On \(\displaystyle 2\le x\le 5\), \(\displaystyle x-5\le 0\) so \(\displaystyle |x-5|=5-x\); on \(\displaystyle 5\le x\le 8\), \(\displaystyle |x-5|=x-5\).\[\int_{2}^{8}|x-5|\,dx=\int_{2}^{5}(5-x)\,dx+\int_{5}^{8}(x-5)\,dx.\]First piece: \[\left[5x-\frac{x^{2}}{2}\right]_{2}^{5}=\left(25-\frac{25}{2}\right)-(10-2)=\frac{25}{2}-8=\frac{9}{2}.\]Second piece: \[\left[\frac{x^{2}}{2}-5x\right]_{5}^{8}=(32-40)-\left(\frac{25}{2}-25\right)=-8+\frac{25}{2}=\frac{9}{2}.\]\[\int_{2}^{8}|x-5|\,dx=\frac{9}{2}+\frac{9}{2}=9.\]

  7. Exercise 7

    ∫01x(1−x)ndx\displaystyle \int_{0}^{1} x(1-x)^{n} d x

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    NCERT’s answer
    \(\displaystyle \frac{1}{(n+1)(n+2)}\)
    Expanding \(\displaystyle (1-x)^{n}\) by the binomial theorem works but is clumsy. Use instead \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=1\), which moves the power onto \(\displaystyle x\).\[I=\int_{0}^{1}x(1-x)^{n}dx=\int_{0}^{1}(1-x)\bigl(1-(1-x)\bigr)^{n}dx=\int_{0}^{1}(1-x)\,x^{n}\,dx.\]Now it is a two-term power integral: \[I=\int_{0}^{1}\left(x^{n}-x^{n+1}\right)dx=\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_{0}^{1}=\frac{1}{n+1}-\frac{1}{n+2}.\]Combining over a common denominator, \[I=\frac{(n+2)-(n+1)}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}.\]Hence \(\displaystyle \displaystyle\int_{0}^{1}x(1-x)^{n}dx=\frac{1}{(n+1)(n+2)}\) (for \(\displaystyle n\) a non-negative integer, and indeed for any \(\displaystyle n>-1\)).

  8. Exercise 8

    ∫0π4log⁡(1+tan⁡x)dx\displaystyle \int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{8} \log 2\)
    Let \(\displaystyle I=\displaystyle\int_{0}^{\pi/4}\log(1+\tan x)\,dx\) and apply \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\frac{\pi}{4}\).The key step is the tangent-difference formula: \[\tan\left(\frac{\pi}{4}-x\right)=\frac{\tan\dfrac{\pi}{4}-\tan x}{1+\tan\dfrac{\pi}{4}\tan x}=\frac{1-\tan x}{1+\tan x}.\]Therefore \[1+\tan\left(\frac{\pi}{4}-x\right)=1+\frac{1-\tan x}{1+\tan x}=\frac{(1+\tan x)+(1-\tan x)}{1+\tan x}=\frac{2}{1+\tan x}.\]So \[I=\int_{0}^{\pi/4}\log\left(\frac{2}{1+\tan x}\right)dx=\int_{0}^{\pi/4}\log 2\,dx-\int_{0}^{\pi/4}\log(1+\tan x)\,dx=\frac{\pi}{4}\log 2-I.\]Hence \(\displaystyle 2I=\dfrac{\pi}{4}\log 2\), i.e. \[I=\frac{\pi}{8}\log 2.\]

  9. Exercise 9

    ∫02x2−xdx\displaystyle \int_{0}^{2} x \sqrt{2-x} d x

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    NCERT’s answer
    \(\displaystyle \frac{16 \sqrt{2}}{15}\)
    Substitution \(\displaystyle t=2-x\) works, and so does the property \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=2\) — take the second, which moves the awkward surd onto \(\displaystyle x\) alone.\[I=\int_{0}^{2}x\sqrt{2-x}\,dx=\int_{0}^{2}(2-x)\sqrt{x}\,dx=\int_{0}^{2}\left(2x^{1/2}-x^{3/2}\right)dx.\]\[I=\left[2\cdot\frac{x^{3/2}}{3/2}-\frac{x^{5/2}}{5/2}\right]_{0}^{2}=\left[\frac{4}{3}x^{3/2}-\frac{2}{5}x^{5/2}\right]_{0}^{2}.\]With \(\displaystyle 2^{3/2}=2\sqrt{2}\) and \(\displaystyle 2^{5/2}=4\sqrt{2}\), \[I=\frac{4}{3}\left(2\sqrt{2}\right)-\frac{2}{5}\left(4\sqrt{2}\right)=\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}=8\sqrt{2}\left(\frac{5-3}{15}\right)=\frac{16\sqrt{2}}{15}.\]Hence \(\displaystyle \displaystyle\int_{0}^{2}x\sqrt{2-x}\,dx=\frac{16\sqrt{2}}{15}\).

  10. Exercise 10

    ∫0π2(2log⁡sin⁡x−log⁡sin⁡2x)dx\displaystyle \int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{2} \log \frac{1}{2}\)
    Simplify the integrand first, using \(\displaystyle \sin 2x=2\sin x\cos x\) and \(\displaystyle \log(abc)=\log a+\log b+\log c\): \[\log\sin 2x=\log 2+\log\sin x+\log\cos x.\]So \[2\log\sin x-\log\sin 2x=2\log\sin x-\log 2-\log\sin x-\log\cos x=\log\sin x-\log\cos x-\log 2.\]Hence \[I=\int_{0}^{\pi/2}\log\sin x\,dx-\int_{0}^{\pi/2}\log\cos x\,dx-\int_{0}^{\pi/2}\log 2\,dx.\]By \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\frac{\pi}{2}\), the substitution \(\displaystyle x\mapsto\frac{\pi}{2}-x\) turns \(\displaystyle \log\sin x\) into \(\displaystyle \log\cos x\), so \[\int_{0}^{\pi/2}\log\sin x\,dx=\int_{0}^{\pi/2}\log\cos x\,dx,\] and the first two integrals cancel exactly. (Each is a convergent improper integral, so the cancellation is legitimate.)What remains is \[I=-\log 2\int_{0}^{\pi/2}dx=-\frac{\pi}{2}\log 2.\]