What is $\int\log(\sin x)~dx$? I know that the value of the integral of $\cot(x)$ is $\log|\sin x|+C$ .
But what about: 
$$\int\log(\sin x)~dx$$
Is there any easy way to find an antiderivative for this? Thanks.

The Maple command $$ int(ln(abs(sin(x))), x); $$ outputs $$i\ln  \left( 2 \right) \ln  \left( {{\rm e}^{ix}} \right) -i\ln 
 \left( {{\rm e}^{ix}} \right) \ln  \left( {\frac {-i \left(  \left( {
{\rm e}^{ix}} \right) ^{2}-1 \right) {\it signum} \left( \sin \left( x
 \right)  \right) }{{{\rm e}^{ix}}}} \right) -1/2\,i \left( \ln 
 \left( {{\rm e}^{ix}} \right)  \right) ^{2}+i\ln  \left( {{\rm e}^{ix
}} \right) \ln  \left( {{\rm e}^{ix}}+1 \right) +i{\it dilog} \left( {
{\rm e}^{ix}}+1 \right) -i{\it dilog} \left( {{\rm e}^{ix}} \right) 
. $$
PS. The Mathematica command $$ Integrate[Log[Abs[Sin[x]]], x, Assumptions -> x > 0\, \&\& x < Pi/2]
 $$ outputs $$Piecewise[{{I [Pi] x - x Log[1 - E^{2 I x}] + x Log[Sin[x]] + 
    1/2 I (x^2 + PolyLog[2, E^{2 I x}]), 
   Sin[x] < 0}}, -x Log[1 - E^{2 I x}] + x Log[Sin[x]] + 
  1/2 I (x^2 + PolyLog[2, E^{2 I x}])] .$$
PPS. Because the formulatian of the question under consideration was edited from $\int \ln|sin(x)|\,dx$ to $\int \ln(sin(x))\,dx$ after my answer, I would like to add that the Maple command $$ int(ln(sin(x)), x) $$ outputs it in a closed form $$-x\ln  \left( 1-{{\rm e}^{2\,ix}} \right) +x\ln  \left( \sin \left( x
 \right)  \right) +1/2\,i{x}^{2}+1/2\,i{\it polylog} \left( 2,{{\rm e}
^{2\,ix}} \right)
  .$$
