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.

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.

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 $$I = -\dfrac{\pi}2 \log2$$