Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$ What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to  
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$

Let $I = \int_0^1 x \tan(\pi x) \log(\sin(\pi x)) dx$. We have
$$I= \int_0^1 \dfrac{x \tan(\pi x)}2 \log(1-\cos^2(\pi x))dx$$
Hence,
$$2I = -\sum_{k=1}^{\infty}\dfrac1k\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx \,\,\,\,\,\,\,\, \spadesuit$$
\begin{align}
\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx & = \int_0^1 x\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx + \int_{1/2}^1 x\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx - \int_0^{1/2} (1-x)\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = 2 \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx - \int_0^{1/2}\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \frac2{\pi^2}\int_0^1 t^{2k-1} \arccos(t) dt - \frac1{\pi}\int_0^1 t^{2k-1} dt\\
& = \frac1{2\pi^{3/2}} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} - \frac1{2 \pi k}
\end{align}
Use this in $\spadesuit$ and the approrpriate Taylor series will give you the answer. I am away now since I have a class to teach. Will add more details if needed sometime later.

Updated
We have
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}2 \dfrac1{(1-x)^{3/2}}$$
Integrate the above once and divide by $x$ to obtain
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}{x(1-x)^{1/2}} + \frac{c_1}x$$
To obtain $c_1$, take $\lim_{x \to 0}$, the left hand side is $\Gamma(1/2)$. For the right hand side limit to even exist, we need $c_1 = -\sqrt{\pi}$. Hence,
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}{x(1-x)^{1/2}} - \frac{\sqrt{\pi}}x$$
Integrate again and divide by $x$ to obtain
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} x^{k-1} = \sqrt{\pi} \left(\dfrac{\log\left(1-\sqrt{1-x} \right) - \log\left(1+\sqrt{1-x} \right)}x \right) - \sqrt{\pi}\frac{\log(x)}x + \frac{c_2}x$$
To obtain $c_2$, take limit $x \to 0$. For the right hand side limit to even exist, we need $c_2 = 2\sqrt{\pi} \log(2)$. Hence,
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} x^{k-1} = \dfrac{2\sqrt{\pi}\log2-2\sqrt{\pi} \log\left(1+\sqrt{1-x}\right)}{x}$$
Integrate again to obtain
\begin{align}
\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^3\Gamma(k)} x^{k} = \frac{\sqrt{\pi}}{2} \left(2\text{Li}_2(1/2(\sqrt{1-x}-1)) - 2\text{Li}_2(1/2(\sqrt{1-x}+1))\\
 + \left(\log(1-\sqrt{1-x}) - \log(1+\sqrt{1-x}) \right)\left(\log(1-\sqrt{1-x}) + \log(1+\sqrt{1-x}) -2\log2\right)\right) - \sqrt{\pi} \frac{\log^2(x)}2 + 2\sqrt{\pi} \log2 \log(x) + c_3
\end{align}
Now plug in $x=1$ in the above to get the sum of the first series and the sum of the second series is $$\sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac{\pi^2}6$$
I shall add more details if needed.
