Computing the integral of $\log(\sin x)$ How to compute the following integral? 
$$\int\log(\sin x)\,dx$$

Motivation: Since $\log(\sin x)'=\cot x$, the antiderivative $\int\log(\sin x)\,dx$ has the nice property $F''(x)=\cot x$. Can we find $F$ explicitly? Failing that, can we find the definite integral over one of intervals where $\log (\sin x)$ is defined?

It seems that the integral can not be evaluated by elemantary functions, but one have to write the solution by using a special function which we have not seen in Calculus $1$. Instead of starting with the product expansion of $\frac{\sin x}{x}$, I saw the following Calculus $1$ trick:
$$\ln\sin x=\ln\sin x+x\cot x-1+2\frac{1-x\cot x}{2}$$
Hence,
$$\int\ln\sin x dx=x\ln\sin x-x+2\pi M\left(\frac{x}{\pi}\right)+c$$
where
$$M(t)=\int_0^t\frac{1-\pi t\cot(\pi t)}{2}dt=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2n+1}t^{2n+1},\hspace{1cm} |t|<1.$$
Then, we have $M(\frac{1}{2})=\frac14-\frac14\ln2$. You may see  WA output.
Hence, $\int_0^{\pi/2}\ln\sin x dx=-\frac\pi 2+2\pi(\frac14-\frac14\ln2)=-\frac\pi 2\ln 2$.
You may argue that for the computation of $M(\frac{1}{2})$, you will need $\int_0^{\pi/2}\ln\sin x dx$, so this is a chicken and egg situation. But, I think there are other methods to compute the values of $M(x)$ involving Gamma, Beta etc. functions.
