Closed form for ${\large\int}_0^\pi\frac{x\,\cos\frac x3}{\sqrt[3]{\sin x}}dx$ I'm trying to find a closed form for the integral below and I found the following conjecture using computer search (and some lucky guesses):
$$\int_0^\pi\frac{x\,\cos\frac x3}{\sqrt[3]{\sin x}}dx\stackrel{\color{#A0A0A0}?}=\frac{\pi\sqrt[3]2}{24}\big(7\pi\sqrt3-9\ln3\big).\tag1$$
Could you please help me to find a proof of it?
 A: As I wrote in my other post, the most complete reference for various integrals is Prudnikov-Brychkov-Marichev. Here it suffices to use the formula 2.5.12.36 from Vol. 1:
$$\int_0^{\pi}\sin^{\mu-1}x\sin bx\,dx=\frac{2^{1-\mu}\pi\Gamma\left(\mu\right) \sin \frac{b\pi}{2}}{\Gamma\left(\frac{\mu+1-b}{2}\right)\Gamma\left(\frac{\mu+1+b}{2}\right)}.\tag{1}$$
(For the proof, make the change of variables $t=e^{2i x}$ in $\displaystyle\int_0^{\pi}e^{ibx}\sin^{\mu-1}x\,dx$ and shrink the resulting integration contour to the branch cut $t\in[0,1]$.) 
Differentiating (1) with respect to $b$, we obtain
\begin{align}
&\int_0^{\pi}x\sin^{\mu-1}x\cos bx\,dx=\\=&\,\frac{2^{-\mu}\pi\Gamma\left(\mu\right) }{\Gamma\left(\frac{\mu+1-b}{2}\right)\Gamma\left(\frac{\mu+1+b}{2}\right)}\left\{\pi\cos\frac{\pi b}{2}+\left[\psi\left(\frac{\mu+1-b}{2}\right)-\psi\left(\frac{\mu+1+b}{2}\right)\right]\sin\frac{b\pi}{2}\right\}.
\end{align}
Setting $b=\frac13$, $\mu=\frac23$ and using that $\psi\left(\frac23\right)-\psi\left(1\right)=\frac{\pi\sqrt{3}-9\ln 3}{6}$ (in fact the digamma function of any rational argument has an elementary expression) yields
$$\int_0^{\pi}\frac{x\cos\frac{x}{3}\,dx}{\sqrt[3]{\sin x}}=2^{-\frac23}\pi\left\{\pi\frac{\sqrt 3}{2}+\frac{\pi\sqrt{3}-9\ln 3}{12}\right\},$$
which is equivalent to the conjectured result.
