Question

Integrate:$\int \frac{sinx}{1+sinx}dx$

Answer 1

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

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

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

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

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

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

$=\int \tan x\sec xdx-\int {\tan }^{2}xdx$

$=\int \tan x\sec xdx-\int {\sec }^{2}xdx+\int dx$

$=\sec x-\tan x+x+c$ where $c$ is the constant of integration.