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
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\pi/2}
\ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x:\ {\large ?}}$

\begin{align}
&\color{#c00000}{\int_{0}^{\pi/2}
\ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x}
\\[3mm]&=\half\int_{-\pi/2}^{\pi/2}
\ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x
\\[3mm]&=\half\int_{-\pi/2}^{\pi/2}
\ln\pars{1-{1 + \cos\pars{2x} \over 2}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta}\,{1 - \cos\pars{2x} \over 2}}}\,\dd x
\\[3mm]&={1 \over 4}\int_{-\pi}^{\pi}
\ln\pars{1 - \half\,\sin^{2}\pars{\alpha}
+ {1 \over 4}\,\sin^{2}\pars{\beta} -\bracks{\half\,\sin^{2}\pars{\alpha}
+ {1 \over 4}\,\sin^{2}\pars{\beta}}\cos\pars{x}}\,\dd x
\end{align}

Let's
$$
\verts{\sin\pars{\alpha}} = {\root{2}\sin\pars{\theta/2} \over a}\,,\quad
\verts{\sin\pars{\beta}} = {2\cos\pars{\theta/2} \over a}\,,\qquad
0 \leq \theta < \pi
$$
such that
\begin{align}
&-\,\half\,\sin^{2}\pars{\alpha} + {1 \over 4}\,\sin^{2}\pars{\beta}
={\cos\pars{\theta} \over a^{2}}\,,\qquad
\half\,\sin^{2}\pars{\alpha} + {1 \over 4}\,\sin^{2}\pars{\beta}
={1 \over a^{2}}
\\[3mm]&\mbox{and}\quad\theta
=2\arctan\pars{\root{2}\,{\verts{\sin\pars{\alpha}} \over \verts{\sin\pars{\beta}}}}
\,,\qquad \verts{a} = {2 \over \root{2\sin^{2}\pars{\alpha} + \sin^{2}\pars{\beta}}}
\end{align}
Note that $\ds{\verts{a} \geq {2\root{3} \over 3} \approx 1.1547 > 1}$.

Then,
  \begin{align}
&\color{#c00000}{\int_{0}^{\pi/2}
\ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x}
\\[3mm]&={1 \over 4}\int_{-\pi}^{\pi}
\ln\pars{1 + {\cos\pars{\theta} \over a^{2}} - {\cos\pars{x} \over a^{2}}}\,\dd x
\\[3mm]&=\color{#c00000}{-\pi\ln\pars{\verts{a}}+{1 \over 4}\int_{-\pi}^{\pi}
\ln\pars{\mu - \cos\pars{x}}\,\dd x}\,,\qquad \mu \equiv a^{2} + \cos\pars{\theta}
\\[3mm]&\mbox{with}\quad
\mu \geq {2\root{3} \over 3} + \cos\pars{\theta} \geq
{2\root{3} - 3 \over 3} \approx 0.1547
\end{align}

The answer is
\begin{align}
&\color{#00f}{\large\int_{0}^{\pi/2}
\ln\pars{1-\cos^{2}\pars{x}\bracks{\sin^{2}\pars{\alpha} - \sin^{2}\pars{\beta} \sin^{2}\pars{x}}}\,\dd x}
\\[3mm]&=\color{#00f}{\large%
-\pi\ln\pars{\verts{a}}
+\half\,\pi\ln\pars{\mu + \root{\mu^{2} - 1} \over 2}}\,,\qquad\mu > 1
\end{align}
A: 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:
$$
I(a,b)=\pi \ln\left( \frac{1+ \sqrt{a} + \sqrt{2\sqrt{a} + a+b + 1}}{4} \right)
$$
A: 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
$$I=\pi \ln\frac{\cos^2\frac\alpha 2+\sqrt{\cos^4 \frac\alpha 2+\cos^2\frac{\beta}{2} \sin^2 \frac{\beta}{2}}}2
$$
