Finding the integral of rational function of sines $$ \int \frac{\sin x}{1+\sin x} \, \mathrm{d}x$$ How do I integrate this? I tried multiplying and dividing by $ (1- \sin x) $.

First, let's simplify things just a bit and write
$$\frac{\sin x}{1+\sin x}=1-\frac{1}{1+\sin x}$$
Then, applying the Wierstrauss substitution $u=\tan(x/2)$ with $\sin x=\frac{2u}{1+u^2}$ and $dx=\frac{2du}{1+u^2}$ reveals that
$$\begin{align}
\int\frac{\sin x}{1+\sin x}dx&=\int \left(1-\frac{1}{1+\sin x}\right)dx\\\\
&=x-\int\frac{2}{1+u^2+2u}du\\\\
&=x-2\int\frac{1}{(u+1)^2}\\\\
&=x+\frac{2}{1+\tan(x/2)}+C
\end{align}$$
Thus, we have 
$$\bbox[5px,border:2px solid #C0A000]{\int \frac{\sin x}{1+\sin x}dx=x+\frac{2}{1+\tan(x/2)}+C}$$
