Question

Evaluate :

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

Answer 1

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

Let $\cos x=t$

$-\sin xdx=dt$

Now, $x=0⟹t=1$ and $x=\frac{\pi }{2}⟹t=0$

Therefore,

$I={{\int }_1}^0\frac{\sin x}{1+t^2}\left(\frac{-dt}{\sin x}\right)$

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

$I=-{[ta{n}^{-1}t{]}_1}^0$

$I=-[ta{n}^{-1}0-ta{n}^{-1}1]=-[0-\frac{\pi }{4}]=\frac{\pi }{4}$