Question

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

Answer 1

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

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

$=\int \frac{x(1-\sin x)}{{\cos }^{2}x}dx$
$I=\int x{\sec }^{2}dx-\int x\sec x\tan xdx.$
Integrate each by parts

$I=x\tan x-\int 1.\tan xdx-x\sec x+\int 1.\sec xdx$

$I=[x\tan x-\log \sec x]-[x\sec x-\log (\sec x+\tan x)]+C$

$I=x\tan x-\log \sec x-x\sec x+\log (\sec x+\tan x)+C$