int_0^pi frac{x sin x}{1+cos ^2 x} d x= | vedclass

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(A) Let $I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^\pi \frac{(\

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

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(A) Let $I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$. $I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. $I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$. $2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. Let $u = \cos x$,then $du = -\sin x dx$. When $x=0, u=1$; when $x=\pi, u=-1$. $2I = \pi \int_1^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^1 \frac{du}{1+u^2}$. $2I = \pi [	an^{-1} u]_{-1}^1 = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$. $I = \frac{\pi^2}{4}$.

$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

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