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$\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x$


$\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x$

Let $\cos x=t \Rightarrow-\sin x d x=d t$

When $x=0, t=1$ and when $x=\frac{\pi}{2}, t=0$

$\Rightarrow \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x=-\int_{1}^{0} \frac{d t}{1+t^{2}}$

$=-\left[\tan ^{-1} t\right]_{1}^{0}$

$=-\left[\tan ^{-1} 0-\tan ^{-1} 1\right]$



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