Question

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

Answer 1

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

Let $\cos x=t$ when $x=0$ $t=1$

$\sin xdx=-dt$, $x=\frac{\pi }{2},t=0$

$I=-{\int }_1^0\frac{1}{1+t^2}dt={\int }_0^1\frac{1}{1+t^2}dt$

$I={\left[{\tan }^{-1}t\right]}_0^1={\tan }^{-1}1-{\tan }^{-1}0$

$I=\frac{\pi }{4}-0=\frac{\pi }{4}$

$[I=\frac{\pi }{4}]$.