power logsine integral $\int_0^\frac{\pi}{2}\ln^n{(\sin{x})}dx$ It is not hard to find that $\int_0^\frac{\pi}{2}\ln{(\sin{x})}dx=-\frac{\pi}{2}\ln2$.
https://socratic.org/questions/how-do-you-prove-that-the-integral-of-ln-sin-x-on-the-interval-0-pi-2-is-converg
Moreover wolfram can compute that 
 $$\int_0^\frac{\pi}{2}\ln^2{(\sin{x})}dx=\frac{\pi^3+3\pi\ln^24}{24}$$
 $$\int_0^\frac{\pi}{2}\ln^3{(\sin{x})}dx=-\frac{\pi}{16}(12\zeta(3)+\ln^34+\pi^2\ln4)$$
and  $$\int_0^\frac{\pi}{2}\ln^4{(\sin{x})}dx=\pi\zeta(3)\ln{8}+\frac{19\pi^5}{480}+\frac{\pi}{2}\ln^42+\frac{\pi^3}{4}\ln^22$$
How to prove these formulas?
In general, How to evaluate the integral $\displaystyle\int_0^\frac{\pi}{2}\ln^n{(\sin{x})}dx$ ?
Thanks in advance.

$\begin{align}J_n=\int_0^{\frac{\pi}{2}}\ln^n(\sin x)\,dx\end{align}$
Perform the change of variable $y=\sin x$,
$\begin{align}J_n=\int_0^1 \frac{\ln^n(x)}{\sqrt{1-x^2}}\,dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}J_n=\frac{1}{2^{n+1}}\int_0^1 \frac{x^{-\frac{1}{2}}\ln^n(x)}{\sqrt{1-x}}\,dx\end{align}$
Consider,
$\begin{align}B_n(s)&=\frac{1}{2^{n+1}}\int_0^1 x^{-\frac{1}{2}+s}(1-x)^{-\frac{1}{2}}\,dx\\
&=\frac{1}{2^{n+1}}\text{B}\left(\frac{1}{2}+s,\frac{1}{2}\right)\\
&=\frac{1}{2^{n+1}}\frac{\Gamma\left(\frac{1}{2}+s\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1+s)}\\
&=\frac{\sqrt{\pi}}{2^{n+1}}\frac{\Gamma\left(\frac{1}{2}+s\right)}{\Gamma(1+s)}
\end{align}$
$\begin{align}\boxed{J_n=\frac{\partial^n B_n(s)}{\partial s^n}\big|_{s=0}}\end{align}$
NB:
$\text{B}(,)$ denotes the Beta Euler function.
Example:
Let $n=1$.
$\begin{align}J_1=\frac{\sqrt{\pi}}{4}\frac{\Gamma^\prime\left(\frac{1}{2}\right)\Gamma(1)-\Gamma\left(\frac{1}{2}\right)\Gamma^\prime(1)}{\Gamma^2(1)}\end{align}$
But,
$\begin{align}\Gamma^\prime\left(\frac{1}{2}\right)=-(\gamma+2\ln 2)\sqrt{\pi}\end{align}$
$\begin{align}\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\end{align}$
$\begin{align}\Gamma\left(1\right)=1\end{align}$
$\begin{align}\Gamma^\prime\left(1\right)=-\gamma\end{align}$
Therefore,
$\begin{align}J_1&=\frac{\sqrt{\pi}}{4}\times \frac{-(\gamma+2\ln 2)\sqrt{\pi}+\gamma\sqrt{\pi}}{1}\\
&=\boxed{-\frac{1}{2}\pi\ln 2}
\end{align}$
Addendum:
To compute $J_n$ one needs to be able to evaluate $\Gamma^{(k)}(1),\Gamma^{(k)}\left(\frac{1}{2}\right)$ for $0\leq k\leq n$.
Fact 1):
$\begin{align} \Gamma^{(m+1)}(z)=\sum_{k=0}^{m} \binom{m}{k}\Gamma^{(k)}(z)\psi^{(m-k)}(z)\end{align}$
(following from the definition of Digamma function and the use of Leibniz-Newton formula)
Fact 2):
For $0<|z-1|<1$,
$\begin{align}\psi(z)=-\gamma+\sum_{k=1}^{\infty}(-1)^{k+1}\zeta(k+1)(z-1)^k\end{align}$
Therefore,
For $k\geq 1$,
$\displaystyle \psi^{(k)}(1)=(-1)^{k+1}k!\zeta(k+1)$
Fact 3):
For $0<\Re(z)<1$,
$\displaystyle \psi(z)+\psi(1-z)=\pi\cot(\pi x)$
(allowing to compute $\displaystyle \psi^{(k)}\left(\frac{1}{2}\right)$)
