Closed form of $ \int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx$ Hello I am trying to solve an incredible integral given by
$$
\int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx=\pi \ln\bigg[\frac{1}{2}\left(\cos^2\alpha +\sqrt{\cos^4 \alpha +\cos^2\frac{\beta}{2} \sin^2 \frac{\beta}{2}}\right)\bigg],\qquad \alpha > \beta >0.
$$
I defined
$$
I\equiv 
\int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx
$$
and using $\cos^2 x=1-\sin^2 x$ but obtained a more complicated expression. Usually it is easier to work with a closed form of Log Sine so I was trying this.  I am not really sure how else to approach this at all.  
The result looks like very nice and similar to something we all know :)
I am looking for real or complex methods to solve this problem.  I am not sure of what substitutions to make but maybe we could work in hyperbolic space

With the shorthands $r=\frac{\sin^2\alpha-\sin^2\beta}{2\sin\beta}$, $s=\frac{\sin^2\alpha+\sin^2\beta}{2\sin\beta}$ and the known integral
$$\int_{0}^{\pi/2} \ln(p + q  \cos^2x)dx
= \pi\ln\frac{\sqrt p+\sqrt{p+q}}2 
$$evaluate
\begin{align}
I=& \int_0^{\pi/2}\ln\left[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\right]dx\\
=& \int_0^{\pi/2}\ln\left[\left(\sqrt{1+r^2}+r+\sin\beta \cos^2x\right)
\left(\sqrt{1+r^2}-r-\sin\beta \cos^2x\right)\right]dx\\
 =& \>\pi\ln\frac{\sqrt{\sqrt{1+r^2}+r}+\sqrt{\sqrt{1+r^2}+s} }2
+\pi\ln\frac{\sqrt{\sqrt{1+r^2}-r}+\sqrt{\sqrt{1+r^2}-s} }2
\end{align}
Combine the two terms with
$\sqrt{a+\sqrt c}+\sqrt{a-\sqrt c} =\sqrt{2a+2\sqrt{a^2-c}}$ to obtain
$$
I=\>\pi\ln \frac{1+\cos\alpha+\sqrt{(1+\cos \alpha)^2+\sin^2\beta}}4$$
which can also be expressed as
$$I=\pi \ln\frac{\cos^2\frac\alpha 2+\sqrt{\cos^4 \frac\alpha 2+\cos^2\frac{\beta}{2} \sin^2 \frac{\beta}{2}}}2
$$
