Question

$\underset{-\mathrm{\pi }/4}{\overset{\mathrm{\pi }/4}{\int }}\frac{1}{1+\sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{1+\sin x}\mathit{d}x.\mathrm{Then}, \\ I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}\mathit{d}x \\ \Rightarrow I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1-\sin x}{1-{\sin }^{2}x}dx \\ \Rightarrow I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1-\sin x}{{\cos }^{2}x}dx \\ \Rightarrow I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\left(\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}\right)dx \\ \Rightarrow I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\left({\sec }^{2}x-\sec x\tan x\right)dx \\ \Rightarrow I={\left[\tan x-\sec x\right]}_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}} \\ \Rightarrow I=\left(1-\sqrt{2}\right)-\left(-1-\sqrt{2}\right) \\ \Rightarrow I=2 \end{gathered}$$