I would like to see how you evaluate
$$S(n)=\int_0^{\pi/2} \log(\sin x)^n\mathrm dx,\qquad n\in\Bbb N_0$$
Here's how I do it.
Start with
$$\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx$$
Use $t=\sin(x)^2$ to see that
$$\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx=\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}\mathrm dt=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$
So
$$\int_0^{\pi/2}\sin(x)^{2a}\mathrm dx=\frac{\sqrt\pi}2\frac{\Gamma(a+\frac12)}{\Gamma(a+1)}$$
Taking $\left(\frac{d}{da}\right)^n$ on both sides
$$2^{n}\int_0^{\pi/2}\sin(x)^{2a}\log(\sin x)^n\mathrm dx=\frac{\sqrt\pi}2\left(\frac{d}{da}\right)^n\,\frac{\Gamma(a+\frac12)}{\Gamma(a+1)}$$
And evaluating at $a=0$,
$$S(n)=\frac{\sqrt{\pi}}{2^{n+1}}\left[\left(\frac{d}{da}\right)^n\,\frac{\Gamma(a+\frac12)}{\Gamma(a+1)}\right]_{a=0}$$
Which is a closed form.
Enjoy!
Best Answer
Enforcing the substitution $x\mapsto \arcsin(x)$ reveals
$$\begin{align} S(n)&=\int_0^{\pi/2}\log^n(\sin(x))\,dx\\\\ &=\int_0^1 \frac{\log^n(x)}{\sqrt{1-x^2}}\,dx\\\\ &=\left.\left(\frac{d^n}{dy^n}\int_0^1 \frac{x^y}{\sqrt{1-x^2}}\,dx\right)\right|_{y=0}\\\\ &=\frac12\left.\left(\frac{d^n}{dy^n}\int_0^1 x^{(y-1)/2}(1-x)^{-1/2}\,dx\right)\right|_{y=0}\\\\ &=\frac12\left.\left(\frac{d^n}{dy^n}B\left(\frac{y+1}{2},\frac12\right)\right)\right|_{y=0}\\\\ &=\frac{\Gamma(1/2)}2\left.\left(\frac{d^n}{dy^n}\left(\frac{\Gamma\left(\frac{y+1}{2}\right)}{\Gamma\left(\frac{y}{2}+1\right)}\right)\right)\right|_{y=0} \end{align}$$