int_0^pi frac{theta sin theta}{1+cos ^2 theta | vedclass

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

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

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

$\int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ is equal to

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