Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$ I'm having trouble computing the integral:
$$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$
I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.
Any suggestions are welcome. Thanks.

We can write the integrand as 
$$\begin{equation*}
\frac{1}{1+\cot x}
\end{equation*}$$
and use the substitution $u=\cot x$. Since $du=-\left( 1+u^{2}\right) dx$ we reduce it to a rational function
$$\begin{equation*}
I:=\int \frac{\sin x}{\sin x+\cos x}dx=-\int \frac{1}{\left( 1+u\right)
\left( u^{2}+1\right) }\,du.
\end{equation*}$$
By expanding into partial fractions and using the identities
$$\begin{eqnarray*}
\cot ^{2}x+1 &=&\csc ^{2}x \\
\arctan \left( \cot x\right)  &=&\frac{\pi }{2}-x \\
\frac{\csc x}{1+\cot x} &=&\frac{1}{\sin x+\cos x}
\end{eqnarray*}$$
we get 
$$\begin{eqnarray*}
I &=&-\frac{1}{2}\int \frac{1}{1+u}-\frac{u-1}{u^{2}+1}\,du \\
&=&-\frac{1}{2}\ln \left\vert 1+u\right\vert +\frac{1}{4}\ln \left(
u^{2}+1\right) -\frac{1}{2}\arctan u +C\\
&=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left(
\cot ^{2}x+1\right) -\frac{1}{2}\arctan \left( \cot x\right) +C \\
&=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left(
\csc ^{2}x\right) +\frac{1}{2}x+\text{ Constant} \\
&=&\frac{1}{2}x-\frac{1}{2}\ln \left\vert \sin x+\cos x\right\vert +\text{
Constant.}
\end{eqnarray*}$$
