$\begin{aligned} & \int_0^{\pi / 2} \frac{\sin x}{1+\cos ^2 \mathrm{x}} \mathrm{dx}=-\int_1^0 \frac{\mathrm{dt}}{1+\mathrm{t}^2}[\text { let } \cos \mathrm{x}=\mathrm{t}] \\ & =-\left[\tan ^{-1} \mathrm{t}\right]_1^0 \\ & =-\left(0-\frac{\pi}{4}\right) \\ & =\frac{\pi}{4}\end{aligned}$