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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