$\begin{aligned} & \text { Let } I=\int_0^\pi \frac{1}{1+\sin x} d x \\ & =\int_0^\pi \frac{1}{1+\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}} d x \\ & =\int_0^\pi \frac{\sec ^2 \frac{x}{2}}{\left(1+\tan \frac{x}{2}\right)^2} d x \\ & \text { Put } \tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t \\ & \therefore \quad I=\int_0^{\infty} \frac{2 d t}{(1+t)^2}=\left[-\frac{2}{1+t}\right]_0^{\infty}=2 \\ & \end{aligned}$