Question

Solve:

${\int }_0^{\frac{\pi }{2}}\frac{\sin xdx}{9+{\cos }^{2}x}$

Answer 1

We have,

${\int }_0^{\frac{\pi }{2}}\frac{\sin xdx}{9+{\cos }^{2}x}$

Let

$\cos x=t$

$\sin xdx=dt$

Therefore,

${\int }_1^0\frac{dt}{9+t^2}$

${\int }_1^0\frac{dt}{t^2+3^2}={{\left[\frac{1}{3}{\tan }^{-1}\frac{t}{3}\right]}_1}^0$

$=[\frac{1}{3}{\tan }^{-1}\frac{0}{3}-\frac{1}{3}{\tan }^{-1}\frac{1}{3}]$

$=[\frac{1}{3}{\tan }^{-1}\tan 0-\frac{1}{3}{\tan }^{-1}\frac{1}{3}]$

$=-\frac{1}{3}{\tan }^{-1}\frac{1}{3}$

Hence, this is the answer.