Log trig integral with radical Show that:
$$\int_{0}^{\Large\frac{\pi}{2}}\frac{\log(\sin\theta)}{\sqrt{1+\sin^{2}\theta}}\ d\theta=-\frac{\Gamma^{2}\left(\dfrac14\right)\sqrt{\dfrac\pi2}}{16},$$  or some other equivalent form. 

Consider the integral
\begin{align}
I = \int_{0}^{\pi/2} \frac{\ln(\sin\theta)}{\sqrt{1+\sin^{2}\theta}} \ d\theta.
\end{align}
By making the substitution $t = \sin\theta$ the integral is reduced to
\begin{align}
I = \int_{0}^{1} \ln(t) \ (1-t^{4})^{-1/2} \ dt.
\end{align}
Making the substitution $u = t^{4}$ yields
\begin{align}
I &= \frac{1}{16} \int_{0}^{1} \ln(u) \ u^{-3/4} \ (1-u)^{-1/2} \ du \\
&= \frac{1}{16} \partial_{\mu} \left. \int_{0}^{1} u^{\mu-1} (1-u)^{-1/2} 
\ du \right|_{\mu=1/4} \\
&= \frac{1}{16} \ B(1/4, 1/2) \left[ \psi(1/4) - \psi(3/4) \right].
\end{align}
Using 
\begin{align}
\psi(1/4) &= - \gamma - \frac{\pi}{2} - 3 \ln(2) \\
\psi(3/4) &= - \gamma + \frac{\pi}{2} - 3 \ln(2) \\
\Gamma(3/4) &= \frac{\sqrt{2} \pi}{\Gamma(1/4)}
\end{align}
leads to 
\begin{align}
I = - \frac{\Gamma^{2}\left( \frac{1}{4} \right)}{16} \sqrt{\frac{\pi}{2}}.
\end{align}
Hence
\begin{align}
\int_{0}^{\pi/2} \frac{\ln(\sin\theta)}{\sqrt{1+\sin^{2}\theta}} \ d\theta
= - \frac{\Gamma^{2}\left( \frac{1}{4} \right)}{16} \sqrt{\frac{\pi}{2}}. 
\end{align}
