On $\int_{-\pi/2}^{\pi/2}\operatorname{Li}_3(\sin x)dx$ and its derivative 
Question 1: How can we prove $$I_3=\int_{-\pi/2}^{\pi/2}\operatorname{Li}_3(\sin x)dx=\frac{\pi}{4}\zeta(3)+\frac{1}{6}\pi\ln^32-\frac{1}{24}\pi^3\ln2?$$
  (where $\displaystyle\operatorname{Li}_s(x):=\sum_{n=1}^\infty\frac{x^n}{n^s}$)
  Question 2: Moreover, can we find a general method (not necessary to be an explicit formula) to find $$I_n=\int_{-\pi/2}^{\pi/2}\operatorname{Li}_n(\sin x)dx$$ where $n\in\mathbb{Z}^+$?

My attempt
Recall the definition of polylogarithm, $$I_3=\int_{-\pi/2}^{\pi/2}\sum_{n=1}^\infty\frac{\sin^nx}{n^3}dx\\
=\sum_{m=1}^\infty\frac1{8m^3}\int_0^{\pi/2}2\sin^{2m}xdx\\
=\sum_{m=1}^\infty\frac{\pi(1/2)_m}{8m^3m!}$$
where $(1/2)_n$ denotes the Pochhammer symbol)
Therefore, we are able to deduce that $$I_3=\frac{1}{16} \pi  \, _5F_4\left(1,1,1,1,\frac{3}{2};2,2,2,2;1\right).$$
Moreover, $I_n$ can be represented to similar $_pF_q$ terms. But, I'm not familiar with $_5F_4$. I have no idea how to turn the hypergeometric term into the answer.
Some equivalent forms of $I_n$:$$I_n=\int_{-1}^1\operatorname{Li}_n(x)\frac{dx}{\sqrt{1-x^2}}$$
$$I_n=\int_0^12^{1-n}\operatorname{Li}_n(x^2)\frac{dx}{\sqrt{1-x^2}}$$
Any answer without using hypergeometric techniques will be highly appreciated.
 A: If $n$ is odd the integral $\int_{-\pi/2}^{\pi/2}(\sin\theta)^{n}\,d\theta$ equals zero, so the computation of $\int_{-\pi/2}^{\pi/2}\text{Li}_3(\sin\theta)\,d\theta$ boils down to the computation of 
$$ \sum_{n\geq 1}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{1}{n^3}=\frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}\left[\frac{1}{4^n}\binom{2n}{n}\right]x^{n-1}\log^2(x)\,dx $$
or (by integration by parts)
$$\int_{0}^{1}\frac{\log^3(x)}{(1-x)^{3/2}}\,dx.$$
This can be easily tackled through the partial derivatives of the Beta function.
A: Not an answer, but close to one...
A rather recent question (Prove that $\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)$) (refering as well to Jack d'Aurizio!) and its answer have striken me because it is exactly $8$ times the result you want to establish !
Thus, I have attempted to transform :
$$J:=\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz \ \ \ \text{equals}\tag{1}$$
by dividing and multiplying it by 16, into :
$$J=16\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{1-(2z-1)^2}}\mathrm dz$$
Making the change of variable $2z-1=x$, and taking into account that $2dz=dx$, we obtain 
$$J=8\int_{-1}^1\frac{\operatorname{Li}_3(\frac{1-x}{2})}{\sqrt{1-x^2}}dx $$
which is very close to $8$ times one of the expressions you give
$$\int_{-1}^1\frac{\operatorname{Li}_3(x)}{\sqrt{1-x^2}}dx,$$
but not the same. How can we reduce this discrepancy ? Is there an adequate relationship for $Li_3$ that can be useful here ?
Remark : I could have awaited (maybe for long...) before to establish that :
$$8 \int_{-1}^1\frac{\operatorname{Li}_3(x)}{\sqrt{1-x^2}}dx=\int_{-1}^1\frac{\operatorname{Li}_3(\frac{1-x}{2})}{\sqrt{1-x^2}}dx \tag{2}$$
(good numerical agreement anyway...)
I think it is more interesting to give it "as it is" as a challenge for others to prove (2) if possible.
A: A hypergeometric representation
$$I_n=\int_{-\pi/2}^{\pi/2}\mathrm{Li}_n(\sin x)dx=\sum_{k\geq0}\frac{1}{(k+1)^n}\int_{-\pi/2}^{\pi/2}\sin(x)^{k+1}dx$$
For even $k+1$, $$i_k=\int_{-\pi/2}^{\pi/2}\sin(x)^{k+1}dx=0$$
and for odd $k+1$,
$$i_k=2\int_0^{\pi/2}\sin(x)^{k+1}dx=\frac{\Gamma(\frac12)\Gamma(1+\frac{k}2)}{\Gamma(1+\frac{k+1}2)}$$
so if $k+1$ is odd then $k=2m$ for some $m\in\Bbb N_0$, giving 
$$I_n=\sum_{m\geq0}\frac{m!\Gamma(1/2)}{(2m+1)^n\Gamma(m+1/2+1)}$$
Set $$t_m=\frac{m!\Gamma(1/2)}{(2m+1)^n\Gamma(m+1/2+1)}\Rightarrow t_0=2$$
So $$\frac{t_{m+1}}{t_m}=\frac{(m+1)^2(m+1/2)^n}{(m+3/2)^{n+1}}\cdot\frac{1}{m+1}$$
Hence (unless I made a mistake)
$$I_n=2\,_{n+2}F_{n+1}\left(1,1,\frac12,\dots,\frac12;\frac32,\dots,\frac32;1\right)$$
