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
 A: $$
\eqalign{
  & \int_{x\, = \,0}^\pi  {\sin (x)\ln (\sin (x))dx}  =  - {1 \over 2}\int_{x\, = \,0}^\pi  {\ln (1 - \cos ^{\,2} (x))d\cos (x)}  =   \cr 
  &  =  - {1 \over 2}\int_{t\, = \,1}^{ - 1} {\left( {\ln (1 - t) + \ln (1 + t)} \right)dt}  = \; \cdots  \cr} 
$$
A: Let $u=\cos(x)$, then
$$
\begin{align}
\int_0^\pi\sin(x)\log(\sin(x))\,\mathrm{d}x
&=\frac12\int_{-1}^1\log\left(1-u^2\right)\,\mathrm{d}u\\
&=\frac12\left(\int_{-1}^1\log(1-u)\,\mathrm{d}u+\int_{-1}^1\log(1+u)\,\mathrm{d}u\right)\\
&=\int_0^2\log(v)\,\mathrm{d}v\\[3pt]
&=\left.v\log(v)-v\right]_0^2\\[9pt]
&=2\log(2)-2\tag{1}
\end{align}
$$

Using the comment by Jack D'Aurizio
$$
\begin{align}
&\int_0^1\sin(\pi x)\log(\Gamma(x))\,\mathrm{d}x\tag{2}\\
&=\int_0^1\sin(\pi x)\log(\Gamma(1-x))\,\mathrm{d}x\tag{3}\\
&=\frac12\int_0^1\sin(\pi x)\log\left(\frac\pi{\sin(\pi x)}\right)\,\mathrm{d}x\tag{4}\\
&=\frac1{2\pi}\int_0^\pi\sin(x)\log(\pi)\,\mathrm{d}x
-\frac1{2\pi}\int_0^\pi\sin(x)\log(\sin(x))\,\mathrm{d}x\tag{5}\\
&=\frac{\log(\pi)}\pi-\frac{\log(2)-1}\pi\tag{6}\\[3pt]
&=\frac{\log(e\pi/2)}\pi\tag{7}
\end{align}
$$
Explanation:
$(3)$: substitute $x\mapsto1-x$
$(4)$: average $(2)$ and $(3)$ and use Euler's Reflection Formula
$(5)$: substitute $x\mapsto x/\pi$
$(6)$: apply $(1)$
$(7)$: algebra
A: 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.
