Closed form of $\int_0^{\frac{\pi}{3}} \log^2(\sin x) \mathrm{d}x$ Inspired by the popularity of these kind of integrals appearing on MSE lately, I actually learned new methods to attack weird integrals by studying the beautiful answers on the similar past questions, so I conjecture $$\int_0^{\frac{\pi}{3}} \log^2(\sin x) \mathrm{d}x$$ has a closed form. How would someone approach this?
PS; To be honest, I'm more interested in the methodology, as on the previous questions I lacked the mathematical background to understand them completely anyway. Which implies I don't want Cleo-like answers.

$\def\B{{\text{B}}}\def\F{{\text{$_2$F$_1$}}}$I find the closed form is really nasty since it's involving the hypergeometric functions. Here is my approach. Let $I$ be the given integral and let $\sin x=\sqrt{t}$, then
\begin{align}
I&=\frac{1}{8}\int_0^{3/4}\frac{\ln^2t}{\sqrt{t}\cdot\sqrt{1-t}}\,dt\\
&=\frac{1}{8}\lim_{u\to\frac{1}{2}}\partial_u^2\,\B\left(\frac{3}{4};u,\frac{1}{2}\right)
\end{align}
where $\text{B}\left(z;u,v\right)$ is the incomplete beta function defined by
\begin{align}
\B\left(z;u,v\right)=\int_0^z t^{u-1}(1-t)^{v-1}\,dt
\end{align}
According to Wolfram MathWorld,  incomplete beta function can represented as hypergeometric function
\begin{align}
\B\left(z;u,v\right)=\frac{z^u}{u}\F\left(u,1-v;1+u;z\right)
\end{align}
Therefore, with help of Wolfram Alpha, we arrive to the following closed-form
\begin{align}
I
=&\,\frac{1}{8}\lim_{u\to\frac{1}{2}}\partial_u^2\,\left(\frac{\left(\frac{3}{4}\right)^u}{u}\F\left(u,\frac{1}{2};1+u;\frac{3}{4}\right)\right)\\
=&\,\frac{\sqrt{3}}{16}\left[\F^{(0,1,0,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)-8\,\F^{(0,0,1,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)\right]+\\
&\,2\left[\F^{(0,0,2,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)+2\,\F^{(0,1,1,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)+\F^{(0,2,0,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)\right]+\\
&\,-\frac{\sqrt{3}}{8}\ln\left(\frac{4}{3}\right)\left[2\,\F^{(0,0,1,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)+\F^{(0,1,0,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)-\frac{8\pi}{3\sqrt{3}}\right]\\
&\,+\frac{\pi}{12}\ln^2\left(\frac{4}{3}\right)+\frac{64\pi}{3\sqrt{3}}
\end{align}
It seems the above closed-form could be simplified into Mr. Mhenny Benghorbal's or Gahawar's answer but unfortunately I am unable to prove it. I wish I could, sorry... (╥﹏╥)
