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

Here is a sketchy calculation. It matches numerically for all values of $a,b$ that I tried.
Abbreviate $a = \cos^2 \alpha$ and $b = \sin^2 \beta$. Then $0 \leq  a,b \leq 1$.
We calculate
$$
I(a,b)= \int _{0} ^{\pi/2} \ln(1-\cos^2 x(1-a-b \sin^2 x)) \,dx
$$
We have 
$$
\partial_aI(a,b)= \int _{0} ^{\pi/2} \frac {\cos^2 x} {1-\cos^2 x(1-a-b \sin^2 x)}  \,dx \\
=
\int _{0} ^{\pi/2} \frac {1} {\tan^2 x+a+b \sin^2 x}  \,dx  \\
=\int _{0} ^{\infty} \frac {1} {(1+t^2)(t^2+a+b \frac {t^2} {1+t^2})}  \,dt  \\
$$
(by substituting $\tan x = t$ and using some trig identities)
$$
= \frac{1}{2} \int _{-\infty} ^{\infty} \frac {1} {t^4+(a+b+1)t^2+a}  \,dt  \\
= \frac{1}{2} \frac {\pi} {\sqrt{a(2 \sqrt{a}+a+b+1)}}
$$
(for example using residues; here it is crucial that $(a+b+1)^2 - 4a^2 \geq 0 $, which is the case because $0 \leq  a,b \leq 1$)
By integrating with respect to $a$ (for example by setting $u = \sqrt{a}$), we hence obtain
$$ 
I(a,b) = \pi \ln{ \left(1 + \sqrt{a} + \sqrt{1 + 2 \sqrt{a} + a + b}  \right) } + C(b)
$$
Here $C(b)$ is an integration constant depending on $b$ only. To fix it, we substitute $a = 1$ in the integral. We obtain, using some trig identities,
$$  
I(1,b) = \int _{0} ^{\pi/2} \ln(1+b \sin^2 x \cos^2 x) \,dx   \\
= \frac{1}{2} \int _{0} ^{\pi} \ln\left(1+  \frac{1}{4} b \sin^2 x\right) \,dx  \\
= \int _{0} ^{\pi/2} \ln\left(1+ \frac{1}{4} b \sin^2 x\right) \,dx \\
= \pi \ln \left(\frac{1+\sqrt{1+b/4}}{2} \right) \\
$$
Here I used a previously calculated integral. On the other hand, we have $I(1,b) = C(b) + \pi \ln(2(1+\sqrt{1+b/4}))$. Comparing yields the value of $C(b)$. Plugging this back in, we get:
$$
I(a,b)=\pi \ln\left( \frac{1+ \sqrt{a} + \sqrt{2\sqrt{a} + a+b + 1}}{4} \right)
$$
