Mathematics

# Show that $\int\limits_0^\pi {\dfrac{{x \sin x}}{{1 + {{\cos }^2}x}}dx = -\dfrac{{{\pi ^2}}}{4}}$

##### SOLUTION

L.H.S

$I=\displaystyle\int_{0}^{\pi }{\dfrac{x\sin x}{1+{{\cos }^{2}}x}}dx$             ……….. (1)

We know that

$\displaystyle\int_{a}^{b}{f\left( x \right)}dx=\displaystyle\int_{a}^{b}{f\left( a+b-x \right)}dx$

Therefore,

$I=\displaystyle\int_{0}^{\pi }{\dfrac{\left( \pi -x \right)\sin \left( \pi -x \right)}{1+{{\cos }^{2}}\left( \pi -x \right)}}dx$

$I=\displaystyle\int_{0}^{\pi }{\dfrac{\left( \pi -x \right)\sin x}{1+{{\cos }^{2}}\left( -x \right)}}dx$

$I=\displaystyle\int_{0}^{\pi }{\dfrac{\left( \pi -x \right)\sin x}{1+{{\cos }^{2}}x}}dx$

$I=\displaystyle\int_{0}^{\pi }{\dfrac{\pi \sin x}{1+{{\cos }^{2}}x}}dx-\displaystyle\int_{0}^{\pi }{\dfrac{x\sin x}{1+{{\cos }^{2}}x}}dx$

$I=\displaystyle\int_{0}^{\pi }{\dfrac{\pi \sin x}{1+{{\cos }^{2}}x}}dx-I$

$2I=\displaystyle\int_{0}^{\pi }{\dfrac{\pi \sin x}{1+{{\cos }^{2}}x}}dx$

$I=\dfrac{1}{2}\displaystyle\int_{0}^{\pi }{\dfrac{\pi \sin x}{1+{{\cos }^{2}}x}}dx$                 ……. (2)

Let $t=\cos x$

$\dfrac{dt}{dx}=-\sin x$

$-dt=\sin xdx$

From equation (2), we get

$I=-\dfrac{1}{2}\displaystyle\int_{1}^{-1}{\dfrac{\pi }{1+{{t}^{2}}}}dt$

$I=-\dfrac{\pi }{2}\left[ {{\tan }^{-1}}\left( t \right) \right]_{1}^{-1}$

$I=-\dfrac{\pi }{2}\left[ {{\tan }^{-1}}\left( -1 \right)-{{\tan }^{-1}}\left( 1 \right) \right]$

$I=-\dfrac{\pi }{2}\left[ \dfrac{3\pi }{4}-\dfrac{\pi }{4} \right]$

$I=-\dfrac{\pi }{2}\left[ \dfrac{\pi }{2} \right]$

$I=-\dfrac{{{\pi }^{2}}}{4}$

Hence, proved.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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