$\int \frac{\sin x}{1+2\sin x}dx$ calculate:
$$\int \frac{\sin x}{1+2\sin x}dx$$
I tried using $\sin x=\dfrac{2u}{u^2+1}$, $u=\tan \dfrac{x}2$ and after Simplification:
$$\int \frac{2u}{u^2+4u+1}×\frac{2}{u^2+1}du$$
and I am not able to calulate that.

Simplify first before substituting,
$$\int \frac{\sin x}{1+2\sin x}dx
=\frac12\int dx -\frac12 \int \frac{1}{1+2\sin x}dx$$
$$=\frac12 x -\frac12 \int  \frac{1+\tan^2\frac x2}{\tan^2\frac x2 +4\tan\frac x2 +1}dx$$
$$=\frac12 x -\int \frac{d(\tan \frac x2)}{\left(\tan^2\frac x2 +2\right)^2 -3}$$
$$=\frac12 x +\frac{1}{\sqrt3}\tanh^{-1}\left( \frac{
\tan\frac x2 +2}{\sqrt3}\right)+C$$
