Mathematics

# $\int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx}$ equals:

$\dfrac{\pi ^2}{4}$

##### SOLUTION
$\begin{array}{l}I = \int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} ......(1)\\I = \int\limits_0^\pi {\dfrac{{(\pi - x)\sin (\pi - x)}}{{1 + {{\cos }^2}(\pi - x)}}dx} \\I = \int\limits_0^\pi {\dfrac{{(\pi - x)\sin x}}{{1 + {{\cos }^2}x}}dx} ......(2)\\Adding\,(1)\,and\,(2)\,we\,get\\2I = \int\limits_0^\pi {\dfrac{{\pi \sin x}}{{1 + {{\cos }^2}x}}dx} \\Let,\cos x = t\\ - \sin xdx = dt\\So,\\2I = \int\limits_1^{ - 1} {\dfrac{{ - \pi dt}}{{1 + {t^2}}}} \\2I = \int\limits_{ - 1}^1 {\dfrac{{\pi dt}}{{1 + {t^2}}}} \\I = \dfrac{\pi }{2}\left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1 = \dfrac{{{\pi ^2}}}{4}\end{array}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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