Evaluation of $\iint_D \frac {\ln(2 - \sin \xi \cos \eta)\sin \xi} {2 - 2\sin \xi \cos \eta + \sin^2 \xi \cos^2 \eta} \mathrm d\xi \; \mathrm d\eta$ Evaluate the following integral:
$$\iint_D \frac {\ln(2 - \sin \xi \cos \eta)\sin \xi} {2 - 2\sin \xi \cos \eta + \sin^2 \xi \cos^2 \eta}  \mathrm d\xi \; \mathrm d\eta$$ where $D = [ 0, \pi/2] \times [ 0, \pi / 2]$.
My observations:


*

*The integral can be rewritten as $$\iint_D \sin \xi f(\sin \xi \cos \eta) \mathrm d \xi \mathrm d \eta$$ where $f(x) = \frac  {\ln (2-x)} {(x-1)^2 + 1}$

*Substituting $\mu = \pi / 2 - \xi$,  $\nu = \pi / 2 - \eta$

*Substituting $x = \sin \eta \cos \xi, y = \sin \eta \sin \xi$

*Expanding in terms of series
Source: The problem appeared in The American Mathematical Monthly
 A: We may exploit the following fact:
$$ \forall a\in\mathbb{N},\qquad \int_{0}^{\pi/2}\int_{0}^{\pi/2}\sin(x)^{a+1}\sin(y)^a\,dx\,dy = \frac{\pi}{2+2a}.\tag{1}$$
It follows that, assuming:
$$ \frac{\log(2-z)}{2-2z+z^2}=\sum_{n\geq 0} c_n z^n \tag{2}$$
our integral equals:
$$ \frac{\pi}{2}\sum_{n\geq 0}\frac{c_n}{n+1} = \frac{\pi}{2}\int_{0}^{1}\frac{\log(2-z)}{2-2z+z^2}\,dz = \frac{\pi}{2}\color{blue}{\int_{0}^{1}\frac{\log(1+z)}{1+z^2}\,dz}.\tag{3}$$
The blue integral is a notorious integral, that can be evaluated by setting $z=\tan\theta$ and exploiting a symmetry, resulting in $\color{blue}{\frac{\pi\log 2}{8}}$. By putting all together,

$$\iint_D \frac {\ln(2 - \sin \xi \cos \eta)\sin \xi} {2 - 2\sin \xi \cos \eta + \sin^2 \xi \cos^2 \eta}  \mathrm d\xi \; \mathrm d\eta = \color{red}{\frac{\pi^2\log 2}{16}}.\tag{4}$$

A: Note that the integral $I$ as given by 
$$I=\int_0^{\pi/2}\int_0^{\pi/2} \frac{\log(2-\sin(\theta)\cos(\phi))}{2-2\sin(\theta)\cos(\phi)+\sin^2(\theta)\cos^2(\phi)}\,\sin(\theta)\,d\theta\,d\phi$$
is the surface integral of $f(x)=\frac{\log(2-x)}{2-2x+x^2}$ over a sphere in the first octant.  Therefore, we can write
$$\begin{align}
I&=\int_0^1 \int_0^{\sqrt{1-x^2}}\frac{\log(2-x)}{2-2x+x^2}\frac{1}{\sqrt{1-x^2-y^2}}\,dy\,dx\\\\
&=\int_0^1 \frac{\log(2-x)}{2-2x+x^2}\left(\int_0^{\sqrt{1-x^2}}\frac{1}{\sqrt{1-x^2-y^2}}\,dy\right)\,dx\\\\
&=\frac{\pi}{2}\int_0^1 \frac{\log(2-x)}{2-2x+x^2}\,dx\\\\
&=\frac{\pi^2\log(2)}{16}
\end{align}$$
where the last equality comes from the analysis in the ensuing note.
NOTE:  EVALUATING $\int_0^1 \frac{\log(2-x)}{2-2x+x^2}\,dx$
$$\begin{align}
\int_0^1 \frac{\log(2-x)}{2-2x+x^2}&=\int_0^1 \frac{\log(1+x)}{1+x^2}\,dx \tag 1\\\\
&=\int_0^{\pi/4} \left(\log(\sin(x)+\cos(x))-\log(\cos(x))\right)\,dx \tag 2\\\\
&=\int_0^{\pi/4} \left(\log(\sqrt{2}\cos(x-\pi/4))-\log(\cos(x))\right )\,dx \tag 3\\\\
&=\frac{\pi \log(2)}{8}+\int_0^{\pi/4} \log(\cos(x-\pi/4))\,dx-\int_0^{\pi/4}\log(\cos(x))\,dx \tag 4\\\\
&=\frac{\pi \log(2)}{8}+\int_0^{\pi/4}\log(\cos(x))\,dx-\int_0^{\pi/4}\log(\cos(x))\,dx \tag 5\\\\
&=\frac{\pi \log(2)}{8}
\end{align}$$
In arriving at $(1)$, we enforced the substitution $x\to 1-x$.
In going from $(1)$ to $(2)$, we enforced the substitution $x\to \tan(x)$.
In going from $(2)$ to $(3)$, we used the identity $A\sin (x)+B\cos(x)=\sqrt{A^2+B^2}\cos\left(x-\arctan2(B,A)\right)$.
In going from $(3)$ to $(4)$, we wrote $\log(\sqrt{2}\cos(x-\pi/4))=\frac{\log(2)}{2}+\log(\cos(x-\pi/4))$ and integrated the constant term over $[0,\pi/4]$.
In going from $(4)$ to $(5)$, we enforced the substitution $x\to \pi/4-x$ and exploited the evenness of the cosine function.
A: Here is the solution from the American Mathematical Monthly:



