Question

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

(a) 0
(b) 1/2
(c) 2
(d) 3/2

Answer 1

(c) 2

$$\begin{gathered} {\int }_0^{\mathrm{\pi }}\frac{1}{1+\sin x}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{1}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{1-\sin x}{1-{\sin }^{2}x}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{1-\sin x}{{\cos }^{2}x}dx \\ ={\int }_0^{\mathrm{\pi }}\left(se{c}^{2}x-secx\tan x\right)dx \\ ={\left[\tan x-secx\right]}_0^{\mathrm{\pi }} \\ =0+1-0+1 \\ =2 \end{gathered}$$