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Evaluate :$\int\limits_0^{\pi}\large\frac{xsin x}{1+cos^2x}dx.$

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$I=\large\int \limits_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}$----(1) using the properties of integrals
$I=\large\int \limits_0^{\pi} \frac{(\pi-x) \sin (\pi -x)}{1+\cos ^2(\pi-x)}dx$
$I=\large\int \limits_0^{\pi} \frac{(\pi-x) \sin x}{1+\cos ^2x}dx$----(2)
Adding (1) and (2)
$2I=\large\pi \int \limits_0^\pi \frac{\sin x}{1+\cos ^2x}dx$
Put $\cos x =t$
$-\sin x dx =dt$
x 0 1
t $\pi$ -1

 

$=>2I=\pi \int \limits_{+1}^{-1} \frac{-dt}{1+t^2}$
$=\pi \int \limits_{-1}^1\frac{dt}{1+t^2}$
$2I=-\pi \bigg[\tan^{-1}t\bigg]_{-1}^1$
$2I=+\pi \bigg[\tan^{-1}(1)-\tan^{-1}(-1)\bigg]$
$2I=\pi\bigg[\large\frac{\pi}{4}+\frac{\pi}{4}\bigg]$
$=\large\frac{\pi^2}{2}$
Therefore $I=\large\frac{\pi^2}{4}$

 

answered Mar 22, 2013 by meena.p
 

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