Integral $\int_0^{\pi/2} \frac{\sin^3 x\log \sin x}{\sqrt{1+\sin^2 x}}dx=\frac{\ln 2 -1}{4}$ Hi I am trying to prove$$
I:=\int_0^{\pi/2} \frac{\sin^3 x\log \sin x}{\sqrt{1+\sin^2 x}}dx=\frac{\ln 2 -1}{4}.
$$
Thanks. 
I am possibly trying to simplify this to obtain something like $2\int_0^{\pi/2} \log \sin x\, dx=-\pi \ln 2 $ since this is easily integrable.  However when I try to simplify the terms 
$$
\frac{\sin^3 x}{\sqrt {1+\sin^2 x}}
$$
I obtain a more complicated integrand.  I am not sure how else to go about this one.  I was trying to possibly write 
$$
I(a)=\int_0^{\pi/2} \frac{\sin^3 a x\log \sin x}{\sqrt{1+\sin^2 x}}dx,\quad I'(a)=\int_0^{\pi/2} \frac{\partial}{\partial a}\left(\frac{\sin^3 ax\log \sin x}{\sqrt{1+\sin^2 x}}\right)\, dx,
$$
but this didn't simplify anything for me.  
I also tried the substitution $y=\sin^2 x$, but couldn't manage to get an integral because of the $\sin 2x$ from the derivative.  
 A: The integral screams for the substitution $u = \sin x$, which transforms it into 
$$
\int_0^1\frac{u^3\log u}{\sqrt{1-u^4}} \ \mathrm{d}u,
$$
another, trickier substitution, $w^2 = 1-u^4$ gives
$$
\frac{1}{2} \int_0^1 \log (1-w^2)^{1/4} \ \mathrm{d}w = {1\over 8} \int_0^1 \log(1+w) +\log (1-w) \ \mathrm{d}w = \frac{\ln 2 -1}{4}.
$$
A: We can derive a more general result:
Consider the integral
\begin{align}
I(a)&=\int_0^{\pi/2}\, \frac{\sin(x)^a}{\sqrt{1+\sin(x)^2}}\, dx\\
&=\int_0^{1}\, \frac{t^a}{\sqrt{1-t^4}}\, dt \tag{subst. $t=\sin(x)$}\\
&=\frac{1}{4}\int_0^{1}\, u^{(a-3)/4} (1-u)^{-1/2}\, du \tag{subst. $t^4=u$}\\
&=\frac{1}{4}\mathrm B\left(\frac{a+1}{4},\frac{1}{2}\right) \tag{1}
\end{align}
The third line represents a form of Beta function.
\begin{align}
\therefore I'(a)&=\int_0^{\pi/2}\, \frac{\sin(x)^a\, \log{\sin{x}}}{\sqrt{1+\sin(x)^2}}\, dx\\
&=\frac{1}{16} \, {\left(\psi_0\left(\frac{a+1}{4} \right)-\psi_0\left(\frac{a+3}{4} \right)  \right)} {\rm B}\left(\frac{a+1}{4},\frac{1}{2}\right) \tag{$\frac{d}{da} (1)$}\\
\implies I'(3)&=\frac{\log{2}-1}{4}
\end{align}
References: Beta function and Polygamma function
A: The change of variables $t=\sin x$ yields
$$
I=\int_0^1\frac{t^3\ln t}{\sqrt{1+t^2}}\frac{dt}{\sqrt{1-t^2}}
=\int_0^1\frac{t^3\ln t}{\sqrt{1-t^4}} dt
$$
Then setting $t^4=u$ simplifies to
$$\eqalign{
I&=\frac{1}{16}\int_0^1\frac{1}{\sqrt{1-u}}\ln u\, du\cr
&=\left[\frac{1}{8}(1-\sqrt{1-u})\ln u\right]_0^1-\frac{1}{8}\int_0^1\frac{1-\sqrt{1-u}}{u} du\cr
&=-\frac{1}{8}\int_0^1\frac{1}{1+\sqrt{1-u}} du;\qquad v\leftarrow1+\sqrt{1-u}\cr
&=\frac{1}{4}\int_1^2\frac{1-v}{v} du=\frac{\ln 2-1}{4}.}
$$
and we are done. $\qquad\square$
