Question

Solve :$\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$

Answer 1

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$

Let $\cos x=t$

$-\sin xdx=dt$

When $x=0,t=1$ and when $x=\frac{\pi }{2},t=0$

Therefore,

$I=\underset{1}{\overset{0}{\int }}\frac{dt}{1+t^2}$

we know that $\int \frac{dx}{x^2+a^2}={\tan }^{-1}\frac{x}{a}+C$

$I=[{\tan }^{-1}x{]}_0^1$

$I=-[{\tan }^{-1}0-{\tan }^{-1}1]=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$