How to evaluate $I=\int\limits_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx$ Prima facie, this integral seems easy to calculate,but alas, this not's case $$I=\int\limits_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx$$
 The numerical value is I=-1.122690024730644497584272...
 How to evaluate this integral?
By against,I find:
 $$I=\int\limits_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(2x)}\,dx=-\frac{\pi^3}{48}$$
 A: The answer is shown to be a fairly simple series and Mathematica identifies that series with a hypergeometric function. I have verified Mathematica's claim and used Mathematica to evaluate the hypergeometric function.

In this answer, it is shown that
$$
\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}
=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\tag{1}
$$
Furthermore, integration by parts shows that
$$
\int_0^1x^n\log(x)\,\mathrm{d}x=-\frac1{(n+1)^2}\tag{2}
$$
Substituting $u=\sin(x)$, we get
$$
\begin{align}
\int_0^{\pi/2}\frac{x\log(\sin(x))}{\sin(x)}\,\mathrm{d}x
&=\int_0^1\frac{\sin^{-1}(u)}{\sqrt{1-u^2}}\frac{\log(u)}u\,\mathrm{d}u\\
&=\int_0^1\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}u^{2k}\log(u)\,\mathrm{d}u\\
&=-\sum_{k=0}^\infty\frac1{(2k+1)^3}\frac{4^k}{\binom{2k}{k}}\tag{3}
\end{align}
$$
Mathematica identifies $(3)$ with the hypergeometric function
$$
\bbox[5px,border:2px solid #A0A0A0]{-{\vphantom{\mathrm{F}}}_4\mathrm{F}_3\left(\color{#C00000}{\tfrac12,\tfrac12},\color{#00A000}{1},\color{#F0A000}{1};\color{#0000FF}{\tfrac32,\tfrac32,\tfrac32};1\right)}\tag{4}
$$
This can easily be verified by looking at the ratios of the terms in $(3)$:
$$
\overbrace{4\vphantom{\frac{()^2}{()}}}^{4^k}\cdot\overbrace{\frac{(k+1)^2}{(2k+2)(2k+1)}}^{1/\binom{2k}{k}}\cdot\overbrace{\frac{(2k+1)^3}{(2k+3)^3}}^{1/(2k+1)^3}
=\frac{\color{#C00000}{(k+\frac12)^2}\color{#00A000}{(k+1)}}{\color{#0000FF}{(k+\frac32)^3}}\cdot1\tag{5}
$$
The orange $1$ cancels the $k!$ in the denominator of the definition of the hypergeometric functions.
Mathematica evaluates $(4)$ as
$$
\bbox[5px,border:2px solid #A0A0A0]{-1.12269002473064449758427221442}\tag{6}
$$
A: The closed form of the integral 
$$\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx.$$
Also note the second integral, that is $$\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(2x)}\,dx$$
can be easily evaluated by combining $2$ integrations by parts (to get again the integral we started with) and beta function.
A: There is a closed-form of your integral.
$$I = \frac{\pi^3}{32} + \frac{\pi}{8}\ln^2 2 - 4\,\Im\left[\operatorname{Li}_3\left(\frac{1+i}2\right)\right],$$
where $\operatorname{Li}_3$ is the trilogarithm function.
We could derive it from robjohn's answer, using Cleo's result here.
You could find a related problem here.
A: By Weierstrass subs, the integral boils down to well-known integrals
$$2 \log (2) \int_0^1 \frac{\arctan(x)}{x} \textrm{d}x+2 \int_0^1 \frac{ \arctan(x)\log (x)}{x} \textrm{d}x-2 \int_0^1 \frac{ \arctan(x)\log (1+x^2)}{x} \textrm{d}x,$$
$$=\frac{\pi^3}{32} + \frac{\pi}{8}\log^2(2) - 4\,\Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}2\right)\biggr\}.$$
End of story
A: Based on manipulation of real functions after the half-angle substitution, the integral can be transformed to an known polylog integral
\begin{align} \int\limits_0^{\pi/2}\frac{x\log(\sin{x})}{\sin x}\,dx
=& -4 \int_0^1 \frac{\ln t\ln(1-t)}{1+t^2} dt \\
= & \>4\>\text{Im} \>\text{Li}_3(1+i) -\frac{3\pi^3}{16}
-\frac\pi4\ln^22
\end{align}
