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.
 A: Mathematica returns the following closed form

$$I=\sqrt{3} \; {}_4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}; \frac{3}{2}, \frac{3}{2},\frac{3}{2} \Bigg| \frac{3}{4}\right) \\
+\frac{\sqrt{3}}{4}\log\frac{16}{9} {}_3F_2\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}; \frac{3}{2}, \frac{3}{2} \Bigg| \frac{3}{4}\right)+\frac{\pi}{12}\log^2 \frac{4}{3} \approx 2.0445154345354680.$$

This is not a particularly nice closed form, as one can imagine. Any attempt to prove this result is likely to be a trying task indeed. Curiously, the first hyper-geometric function is somewhat similar to one that I had come across some time ago.
Additionally, through the substitution $x = \arcsin t$, one can find that 

$$I = \frac{1}{2}\frac{d^2}{ds^2} \mathrm{B}\left(\frac{3}{4}, \frac{s}{2}, \frac{1}{2}\right)\Bigg|_{s=1}$$

where $\mathrm{B}(x; \, a,b)$ is the incomplete beta function.
A: $\def\B{{\text{B}}}\def\F{{\text{$_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... (╥﹏╥)
A: Here is a closed form.

$$ I = \frac{\sqrt {3}}{2}\, \left( -2+\ln  \left( 3 \right) -2\,\ln  \left( 2 \right)\right) \,
{\mbox{$_5$F$_4$}\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\alpha;\,\frac{3}{2},\frac{3}{2},\frac{3}{2},\beta;\,\frac{3}{4}\right)}\sim 2.044515433$$

where $ {\mbox{$_5$F$_4$}\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\alpha;\,\frac{3}{2},\frac{3}{2},\frac{3}{2},\beta;\,\frac{3}{4}\right)} $ is the generalized hypergeometric function and $\alpha$, $\beta$ are given by

$$ \alpha = {\frac {-2+3\,\ln  \left( 3 \right) -6\,\ln  \left( 2 \right) }{-4\,\ln  \left( 2 \right) +2\,\ln  \left( 3 \right) }}\\ \\
\beta = {\frac {-2+\ln  \left( 3 \right) -2\,\ln  \left( 2 \right) }{-4\,\ln  \left( 2 \right) +2\,\ln  \left( 3 \right) }} $$

