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.
 A: No such antiderivative can be written using elementary functions.
Rather than $\log (\sin x)$, let us consider the equivalently difficult $\log (\cos x)$.
Using integration by parts, we find:
$$\int \log(\cos x) = x\log(\cos x) + \int x \tan x dx$$
Now, finding the antiderivative means tackling $x \tan x$, and this latter expression has no antiderivative with elementary functions. This result follows from a theorem of Liouville (see, e.g., here) and the specifics of the argument for $x \tan x$ can be found here.
A: The indefinite integral cannot be expressed in terms of elementary functions. However the definite integral from $0$ to $\pi/2$ (or $\pi$) can be avalauted as shown below. $$I=\int_0^{\pi/2} \log(\sin(x)) dx = - \dfrac{\pi \log2}2$$
The above can be evaluated as follows. We have
$$I=\underbrace{\int_0^{\pi/2} \log(\sin(x)) dx = -\int_{\pi/2}^0 \log(\cos(y)) dy}_{y = \pi/2-x} = \int_0^{\pi/2} \log(\cos(x))dx$$
Hence,
$$I+I = \int_0^{\pi/2} \log(\sin(x)) dx + \int_0^{\pi/2} \log(\cos(x)) dx = \int_0^{\pi/2} \log(\sin(x) \cos(x))dx$$
Hence,
$$2I = \int_0^{\pi/2} \log(\sin(2x))dx - \dfrac{\pi}2 \log2 = \dfrac12\int_0^{\pi} \log(\sin(x)) dx - \dfrac{\pi}2 \log2 = I - \dfrac{\pi}2 \log2$$
Hence, we get that
$$I = -\dfrac{\pi}2 \log2$$
A: using $$\ln \sin x =-\ln 2-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}, \ x \in [0,\pi].$$
$$\int \ln(\sin x)dx=-\ln(2)\int dx-\sum^{\infty}_{n=1}\frac{1}{n}\int (\cos 2nx)dx$$
$$=-\ln(2)\cdot x-\sum^{\infty}_{n=1}\frac{\sin(2nx)}{2n^2}+\mathcal{C}$$
A: 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 $$-x\ln  \left( 1-{{\rm e}^{2\,ix}} \right) +x\ln  \left( \sin \left( x
 \right)  \right) +1/2\,i{x}^{2}+1/2\,i{\it polylog} \left( 2,{{\rm e}
^{2\,ix}} \right)
  .$$
