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

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(C) Let $I = \int_0^{\pi / 2} \frac{\pi \sin x}{1+\cos ^2 x} d x$.Substitute $\cos x = t$, then $-\sin x d x = d t$, or $\sin x d x = -d t$.When $x =

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

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(C) Let $I = \int_0^{\pi / 2} \frac{\pi \sin x}{1+\cos ^2 x} d x$.Substitute $\cos x = t$, then $-\sin x d x = d t$, or $\sin x d x = -d t$.When $x = 0$, $t = \cos(0) = 1$.When $x = \pi / 2$, $t = \cos(\pi / 2) = 0$.The integral becomes:$I = \pi \int_1^0 \frac{-d t}{1+t^2} = \pi \int_0^1 \frac{d t}{1+t^2}$.Using the standard integral $\int \frac{1}{1+t^2} d t = 	an^{-1}(t)$:$I = \pi [	an^{-1}(t)]_0^1 = \pi (	an^{-1}(1) - 	an^{-1}(0))$.Since $	an^{-1}(1) = \pi / 4$ and $	an^{-1}(0) = 0$:$I = \pi (\pi / 4 - 0) = \frac{\pi^2}{4}$.

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

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