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$\int_0^{\pi / 2} \frac{\pi \sin x}{1+\cos ^2 x} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{\pi^2}{4}$
Let $I=\int_0^{\pi / 2} \frac{\pi \sin x}{1+\cos ^2 x} d x$
Put $\cos x=t \Rightarrow-\sin x d x=d t$
$\begin{aligned} I & =\pi \int_1^0 \frac{-d t}{1+t^2}=\pi \int_0^1 \frac{1}{1+t^2} d t=\pi\left[\tan ^{-1}(t)\right]_0^1 \\ & =\pi\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right]=\pi^2 / 4\end{aligned}$
Put $\cos x=t \Rightarrow-\sin x d x=d t$
$\begin{aligned} I & =\pi \int_1^0 \frac{-d t}{1+t^2}=\pi \int_0^1 \frac{1}{1+t^2} d t=\pi\left[\tan ^{-1}(t)\right]_0^1 \\ & =\pi\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right]=\pi^2 / 4\end{aligned}$
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