Interesting log sine integrals $\int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2} \right)dx= \frac{7\pi^3}{108}$ Show that
$$\begin{aligned} \int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2}
 \right)dx &= \frac{7\pi^3}{108} \\ \int_0^{\pi/3}x\log^2
 \left(2\sin\frac{x}{2} \right)dx &= \frac{17\pi^4}{6480}\end{aligned}$$


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*I can solve $\displaystyle \int_0^\pi \log^2 \left(2\sin \frac{x}{2}
 \right)dx $ but I don't know what to do if the limits are from $0$ to $\pi/3$. 

*I have no idea what to do if the integrand contains an $x$.

*I feel that the Polylogarithm function will be involved however I don't know how it can be implemented here.


It would be really great if someone could take the initiative to prove these.

The best way I see to do the simpler integral (the one without the $x$ in front) is to substitute $u=2 \sin{(x/2)}$ and expand the resulting integrand in a series.  To wit, upon doing the substitution, we get
$$\int_0^{\pi/3} dx \: \log^2{[2 \sin{(x/2)}]} = \int_0^1 du \frac{\log^2{u}}{\sqrt{1-u^2/4}}$$
Note that
$$\frac{1}{\sqrt{1-u^2/4}} = \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} u^{2 k}$$
Then the integral becomes
$$\sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} \int_0^1 du\: u^{2 k} \log^2{u}$$
The integral on the right may be done through integration by parts; the process is very interesting, but I leave it to the reader to get the nifty result that
$$\int_0^1 du\: u^{2 k} \log^2{u}=\frac{2}{(2 k+1)^3}$$
Therefore the evaluation of the integral becomes an evaluation of the following sum:
$$2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k}\frac{1}{(2 k+1)^3}$$
To evaluate this sum, define
$$f(z) = 2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k}\frac{z^{2 k+1}}{(2 k+1)^3}$$
The desired integral is $f(1)$.  To derive an equation for $f$, recall the binomial series above:
$$\frac{d}{dz}\left [z \frac{d}{dz} \left [ z \frac{d}{dz} f(z)\right ] \right ] = 2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} z^{2 k} = \frac{2}{\sqrt{1-z^2/4}}$$
The resulting integrations are elementary except for the last one, which results in a nasty generalized hypergeometric function, something like
$$2 _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{1}{4}\right)$$
which in fact does numerically check out.
