${\int }_0^{\mathrm{\pi }/2} \frac{\sin \mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}$ is equal to

Let $\mathrm{I}={\int }_0^{\mathrm{\pi }/2} \frac{\sin \mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}$

put $\cos \mathrm{x}=\mathrm{t}\Rightarrow -\sin \mathrm{xdx}=\mathrm{dt}\Rightarrow \mathrm{dx}=-\frac{\mathrm{dt}}{\sin \mathrm{x}}$

For limit when $\mathrm{x}=0\Rightarrow \mathrm{t}=\cos 0=1[\because \mathrm{t}=\cos \mathrm{x}]$

and when $\mathrm{x}=\frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{t}=\cos \frac{\mathrm{\pi }}{2}=0$

$\begin{array}{l}\therefore \mathrm{I}={\int }_1^0 \frac{\sin \mathrm{x}}{1+{\mathrm{t}}^2}\cdot \left(\frac{\mathrm{dt}}{-\sin \mathrm{x}}\right).\\ =-{\int }_1^0 \frac{1}{1+{\mathrm{t}}^2}\mathrm{dt}={\int }_0^1 \frac{1}{1+{\mathrm{t}}^2}\mathrm{dt}\\ ={\left[\frac{1}{1}{\tan }^{-1}\left(\frac{\mathrm{t}}{1}\right)\right]}_1^0\left[\because \int \frac{1}{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{\mathrm{a}}{\tan }^{-1}\frac{\mathrm{x}}{\mathrm{a}}\right]\\ =-\left(0-\frac{\mathrm{\pi }}{4}\right)=\frac{\mathrm{\pi }}{4}\end{array}$