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
$\begin{eqnarray*} I &=&-\frac{1}{2}\int \frac{1}{1+u}-\frac{u-1}{u^{2}+1}\,du \\ &=&-\frac{1}{2}\ln \left\vert 1+u\right\vert +\frac{1}{4}\ln \left( u^{2}+1\right) -\frac{1}{2}\arctan u +C\\ &=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left( \cot ^{2}x+1\right) -\frac{1}{2}\arctan \left( \cot x\right) +C \\ &=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left( \csc ^{2}x\right) +\frac{1}{2}x+\text{ Constant} \\ &=&\frac{1}{2}x-\frac{1}{2}\ln \left\vert \sin x+\cos x\right\vert +\text{ Constant.} \end{eqnarray*}$