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

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

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

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(D) Let $I = \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx \qquad ....(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:$I = \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx \qquad ....(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^\pi \frac{\pi \sin x}{1+\cos^2 x} dx$$I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$ and when $x=\pi, t=-1$.$I = \frac{\pi}{2} \int_1^{-1} \frac{-dt}{1+t^2} = \frac{\pi}{2} \int_{-1}^1 \frac{dt}{1+t^2}$$I = \frac{\pi}{2} [	an^{-1}(t)]_{-1}^1 = \frac{\pi}{2} [	an^{-1}(1) - 	an^{-1}(-1)]$$I = \frac{\pi}{2} [\frac{\pi}{4} - (-\frac{\pi}{4})] = \frac{\pi}{2} [\frac{\pi}{2}] = \frac{\pi^2}{4}$

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

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