Special Gamma function integral I'm trying to evaluate this integral $$\int_{0}^{1} \sin (\pi x)\ln (\Gamma (x)) dx$$
and I got to the point, when I need to find 
$\displaystyle \int_{0}^{\pi } \sin (x)\ln (\sin (x)) dx$
but everything I tried just failed,or either I was not able to put in the borders . Could you please help me.
Thanks

Let we put everything together.
$$ I = \int_{0}^{1}\sin(\pi x)\log\Gamma(x)\,dx = \int_{0}^{1}\sin(\pi z)\log\Gamma(1-z)\,dz \tag{1}$$
leads to:
$$ I = \frac{1}{2}\int_{0}^{1}\sin(\pi x)\log\left(\Gamma(x)\,\Gamma(1-x)\right)\,dx \tag{2}$$
but $\Gamma(x)\,\Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$, hence:
$$ I = \frac{\log \pi}{\pi}-\frac{1}{\pi}\int_{0}^{\pi/2}\sin(x)\log\sin(x)\,dx\tag{3} $$
or, with a change of variable and integration by parts:
$$ I = \frac{\log \pi}{\pi}-\frac{1}{\pi}\int_{0}^{1}\frac{x\log x}{\sqrt{1-x^2}}\,dx = \frac{\log \pi}{\pi}+\frac{1}{\pi}\int_{0}^{1}\frac{1-\sqrt{1-x^2}}{x}\,dx\tag{4}$$
so:

$$ \int_{0}^{1}\sin(\pi x)\log\Gamma(x)\,dx = \color{red}{\frac{1}{\pi}\left(1+\log\frac{\pi}{2}\right)}\tag{5}$$

since a primitive for $\frac{1-\sqrt{1-x^2}}{x}=\frac{x}{1+\sqrt{1-x^2}}$ is given by $\log(1+\sqrt{1-x^2})-\sqrt{1-x^2}$.
By using the reflection formula, $(5)$ can be seen as a consequence of Raabe's formula, too.
It also follows from Kummer's Fourier series expansion.
