Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $ Evaluate

$$\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta $$

Several days ago,I found this interesting integral from a paper about generalized log-sine integrals,but I can't remember the title of it. The answer of the integral is

\begin{align*}
-\mathrm{Ls}_{7}^{\left ( 3 \right )}\left ( \pi  \right)&=\frac{9}{35}\log^72+\frac{4}{5}\pi ^{2} \log^52+9\zeta \left ( 3 \right )\log^42-\frac{31}{30}\pi ^{4}\log^32\\
&-\left [ 72\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{9}{8}\zeta \left ( 5 \right )-\frac{51}{4}\pi ^{2}\zeta \left ( 3 \right ) \right ]\log^22\\
&+\left [ 72\mathrm{Li}_{5,1}\left ( \frac{1}{2} \right )-216\mathrm{Li}_6\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_4\left ( \frac{1}{2} \right ) \right ]\log2+72\mathrm{Li}_{6,1}\left ( \frac{1}{2} \right )\\
&-216\mathrm{Li}_7\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{1161}{32}\zeta \left ( 7 \right )-\frac{375}{32}\pi ^{2}\zeta \left ( 5 \right )+\frac{1}{10}\pi ^{4}\zeta \left ( 3 \right )
\end{align*}
  where 
  $$\mathrm{Ls}_n^{\left ( k \right )}\left ( \alpha  \right ):=-\int_{0}^{\alpha }\theta ^{k}\log^{n-1-k}\left | 2\sin\frac{\theta }{2} \right |\mathrm{d}\theta $$
  is the generalized log-sine integral and
  $$\mathrm{Li}_{\lambda ,1}\left ( z \right )=\sum_{k=1}^{\infty }\frac{z^{k}}{k^{\lambda }}\sum_{j=1}^{k-1}\frac{1}{j}$$
  is the multiple polylogarithm.


I found a beautiful way to solve the integrals below
$$\int_{0}^{\frac{\pi }{2}}t^{2n}\log^{m}\left ( 2\cos t  \right )\mathrm{d}t $$
Let's consider
$$\mathcal{I}\left ( x,y \right )=\int_{0}^{\frac{\pi }{2}}\cos\left ( xt \right )\left ( 2\cos t \right )^{y}\mathrm{d}t$$
By using Gamma function,the integral become
$$\mathcal{I}\left ( x,y \right )=\frac{\pi \, \Gamma \left ( y+1 \right )}{2\Gamma \left ( \dfrac{x+y+2}{2} \right )\Gamma \left ( \dfrac{y-x+2}{2} \right )}$$ 
Then we can get
$$\mathcal{I}\left ( x,y \right )=\frac{\pi }{2}\exp\left ( \sum_{k=2}^{\infty }\frac{\left ( -1 \right )^{k}}{k\cdot 2^{k}}\zeta \left ( k \right )\left [ \left ( 2y \right )^{k}-\left ( y-x \right )^{k}-\left ( x+y \right )^{k} \right ] \right )$$
On the other hand,using taylor series 
$$\mathcal{I}\left ( x,y \right )=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{\left ( 2n \right )!}x^{2n}\sum_{m=0}^{\infty }\frac{y^{m}}{m!}\int_{0}^{\frac{\pi }{2}}t^{2n}\log^m\left ( 2\cos t \right )\mathrm{d}t$$
So,the comparison of coefficient shows the answer.For example
$$\int_{0}^{\frac{\pi }{2}}t^{2}\log^2\left ( 2\cos t \right )\mathrm{d}t=4\cdot \frac{\pi }{2}\left [ \frac{12}{4\cdot 16} \zeta \left ( 4 \right )+\frac{1}{2}\frac{8}{2^{2}\cdot 4^{2}}\zeta \left ( 2 \right )^{2}\right ]=\frac{11}{1440}\pi ^{5}$$

I wonder can we use the same way to prove the integral in the beginning,if not,is there another way to handle it? 

I think you can get from the paper( Jonathan M. Borwein and
and Armin 2013） reslut Log-sine evaluations of Mahler measures Theorem 2.6 use this identity
$$-\sum_{n,k\ge 0}Ls^{(k)}_{n+k+1}(\pi)\dfrac{\lambda ^n}{n!}\cdot\dfrac{i\mu)^k}{k!}=i\sum_{n\ge 0}(-1)^n\binom{\lambda}{n}\dfrac{e^{i\pi\frac{\lambda}{2}}-(-1)^ne^{i\pi\mu}}{\mu-\dfrac{\lambda}{2}+n}$$
then
$$\int_{0}^{\pi}\theta^3\log^3{\left(2\sin{\dfrac{\theta}{2}}\right)}d\theta=-Ls_{7}^{(3)}=\dfrac{d^3}{d\mu^3}\dfrac{d^3}{d\lambda^3}\sum_{n\ge 0}\binom{n}{\lambda}\dfrac{(-1)^ne^{i\pi\frac{\lambda}{2}}-e^{i\pi\mu}}{\mu-\dfrac{\lambda}{2}+n}=6\pi^2\lambda_{5}\left(\dfrac{1}{2}\right)+36Li_{5,1,1}(-1)-\pi^4\zeta{(3)}-\dfrac{759}{32}\pi^2\zeta{(5)}-\dfrac{45}{32}\zeta{(7)}$$
where 
$$\lambda_{n}(x)=(n-2)!\sum_{k=0}^{n-2}\dfrac{(-1)^k}{k!}Li_{n-k}(x)\log^k|x|+\dfrac{(-1)^n}{n}\log^{n}|x|$$
