Question

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

Answer 1

${\int }_0^{2\pi }\frac{\cos x}{1+{\sin }^{2}x}dxLet\sin x=t\Rightarrow \cos xdx=dt\because \int \frac{\cos x}{1+{\sin }^{2}x}dx=\int \frac{dt}{1+t^2}={\tan }^{-1}t={\tan }^{-1}\left(\sin x\right)\therefore {\int }_0^{2\pi }\frac{\cos x}{1+{\sin }^{2}x}dx={\left[{\tan }^{-1}\right(\sin x\left)\right]}_0^{2\pi }={\tan }^{-1}\left(\sin 2\pi \right)-{\tan }^{-1}\left(\sin 0\right)=0-0=0$