Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{1}{1+\sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\mathrm{\pi }}\frac{1}{1+\sin x}\mathit{}\mathit{d}x.\mathrm{Then}, \\ \mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{1-\sin \mathrm{x}}{\left(1+\sin \mathrm{x}\right)\left(1-\sin \mathrm{x}\right)}\mathit{}\mathit{d}\mathrm{x} \\ \Rightarrow I={\int }_0^{\mathrm{\pi }}\frac{1-\sin x}{1-{\sin }^{2}x}dx \\ \Rightarrow I={\int }_0^{\mathrm{\pi }}\frac{1-\sin x}{{\cos }^{2}x}dx\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right] \\ \Rightarrow I={\int }_0^{\mathrm{\pi }}{\sec }^{2}x-\sec x\tan xdx \\ \Rightarrow I={\left[\tan x-\sec x\right]}_0^{\mathrm{\pi }} \\ \Rightarrow I=\left(\tan \pi -\sec \mathrm{\pi }\right)-\left(\tan 0-\sec 0\right) \\ \Rightarrow I=0+1-\left(0-1\right) \\ \Rightarrow I=1+1 \\ \Rightarrow I=2 \end{gathered}$$