Question

Integrate $\int \frac{1}{1+\sin x}dx$

Answer 1

$\int \frac{1}{1+\sin x}dx$

$=\int \frac{1-\sin x}{\left(1+\sin x\right)\left(1-\sin x\right)}dx$

$=\int \frac{1-\sin x}{1-{\sin }^{2}x}dx$

$=\int \frac{1-\sin x}{{\cos }^{2}x}dx$

$=\int \frac{1}{{\cos }^2}dx-\int \frac{\sin x}{{\cos }^{2}x}dx$

$=\int {\sec }^{2}x-\int \sec x\mathrm{tanxdx}$

$=\tan x-\sec x+c$