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.

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.
