I am trying to integrate a special case of the log sine integral $\rm{Ls}_n(\sigma)$ at $\sigma=\pi/3$ : $$ \rm{Ls}_{n}\big(\tfrac{\pi}{3}\big)=-\int_0^{\pi/3}\bigg[\ln\big(2\sin\tfrac{\theta}{2}\big)\bigg]^{n-1}\mathrm d\theta $$ where $n$ is a non-negative integer. This problem is strongly related to the hypergeometric form of the Log Sine integral.
The closed form is rather simple, although I am having trouble computing it. We can use standard log rules on the inside expression, although I am not sure how this will help us...Thanks
$$\text{Consider }\int_0^\frac{\pi}{3}\left(2\sin\frac{\theta}{2}\right)^a~d\theta~,$$
$$\int_0^\frac{\pi}{3}\left(2\sin\frac{\theta}{2}\right)^a~d\theta$$
$$=2^a\int_0^\frac{\pi}{3}\sin^a\frac{\theta}{2}d\theta$$
$$=2^{a+1}\int_0^\frac{\pi}{6}\sin^a\theta~d\theta$$
$$=2^{a+1}\int_0^\frac{1}{2}x^a~d(\sin^{-1}x)$$
$$=2^{a+1}\int_0^\frac{1}{2}\dfrac{x^a}{\sqrt{1-x^2}}dx$$
$$=2^{a+1}\int_0^\frac{1}{4}\dfrac{x^\frac{a}{2}}{\sqrt{1-x}}d\left(x^\frac{1}{2}\right)$$
$$=2^a\int_0^\frac{1}{4}\dfrac{x^\frac{a-1}{2}}{\sqrt{1-x}}dx$$
$$=2^aB\left(\dfrac{1}{4};\dfrac{a+1}{2},\dfrac{1}{2}\right)$$
$$\therefore\int_0^\frac{\pi}{3}\ln^n\left(2\sin\frac{\theta}{2}\right)d\theta=\dfrac{d^n}{da^n}\left(2^aB\left(\dfrac{1}{4};\dfrac{a+1}{2},\dfrac{1}{2}\right)\right)(a=0)$$