Answer
\[\frac{π}{4}\]
Work Step by Step
Let \[I=\int_{0}^{\frac{π}{2}}\frac{\sin x}{1+\cos ^2 x}\:dx\]
Put $t=\cos x\;\;\Rightarrow \;\;dt=-\sin x\:dx$
Also at $x=0\;\;\Rightarrow \;\; t=1$
and at $x=\frac{π}{2}\;\;\Rightarrow \;\; t=0$
\[\Rightarrow I=-\int_{1}^{0}\frac{1}{1+t^2}\:dt\]
\[I=-\left[\tan^{-1}t\right]_{1}^{0}\]
\[I=\tan^{-1}(1)-\tan^{-1}(0)\]
\[\Rightarrow I=\frac{π}{4}\]
Hence \[I=\frac{π}{4}\]
