$\displaystyle \int{\frac{x+sin(x)}{1+cos(x)}}dx$

My attempt so far

$\displaystyle \int{\frac{(x+sin(x))(1-cos(x))}{(1+cos(x))(1-cos(x))}}dx$ =

$\displaystyle \int{\frac{x-x*cos(x)+sin(x)-cos(x)sin(x)}{sin^2(x)}}dx$ =

$\displaystyle -x*cotan(x)+ln(sin(x))-x*cosec(x)-\frac{cos(x)}{sin^2(x)} + \int{\frac{1}{sin(x)}}dx-ln(sin(x))$ =

$\displaystyle -x*cotan(x)-x*cosec(x)-\frac{cos(x)}{sin^2(x)} + \int{\frac{1}{sin(x)}}dx$

I should get $\displaystyle \frac{x(1-cos(x))}{sin(x)} = \frac{x}{sin(x)}-x*cotan(x)$ pretending that I don't know how to solve $\displaystyle \int\frac{1}{sin(x)}dx$ so apparantly somehow I should be able to paritally integrate $\displaystyle \int\frac{x*cos(x)}{sin^2(x)}dx$ in such a way that I get $\displaystyle \int\frac{-1}{sin(x)}dx$ and so get rid of that integral.