Huge Triple Integral Does there exist an easy method for proving that
$$
\int_1^2\int_1^2\int_1^2\arcsin\left(\frac{\sqrt{x^2-1}+\sqrt{y^2-1}+\sqrt{z^2-1}-\sqrt{x^2-1}\sqrt{y^2-1}\sqrt{z^2-1}}{xyz}\right)\,dx\,dy\,dz
=3\log(2+\sqrt{3})-\pi.
$$
Can someone give me an easy method?
 A: The radical $\sqrt{x^2-1}$ is one leg of a right triangle whose other leg is $1$ and whose hypotenuse is $x$.  Hence
$$\sqrt{x^2-1}=\tan\alpha\text{ and }x=\sec\alpha$$
where you'll see which angle $\alpha$ is if you draw the picture. Similarly
$$
\begin{align}
\sqrt{y^2-1} = \tan\beta, & &y = \sec\beta, \\[6pt]
\sqrt{z^2-1} = \tan\gamma, & &z = \sec\gamma.
\end{align}
$$
Now
$$
\tan(\alpha+\beta+\gamma)=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma},
$$
and
$$
\sec(\alpha+\beta+\gamma)=\frac{\sec\alpha\sec\beta\sec\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}.
$$
So the function in your integral is
$$
\arcsin\left(\frac{\tan(\alpha+\beta+\gamma)}{\sec(\alpha+\beta+\gamma)}\right) = \arcsin(\sin(\alpha+\beta+\gamma))=\alpha+\beta+\gamma.
$$
Then you have
$$
dx = \sec\alpha\tan\alpha\,d\alpha
$$
and similarly for $dy$ and $dz$.  And
$$
\int_1^2 \cdots\cdots dx
$$
becomes
$$
\int_0^{\pi/3} \cdots\cdots (\sec\alpha\tan\alpha\,d\alpha)
$$
since $\sec0=1$ and $\sec\dfrac\pi3=2$.
Then make it a sum of three integrals, all with the same numerical value, one of which is
$$
\iiint \alpha \sec\alpha \tan\alpha \sec\beta \tan\beta \sec\gamma \tan\gamma \,d\alpha \,d\beta \,d\gamma
$$
$$
= \int_0^{\pi/3} \alpha \sec\alpha \tan\alpha\,d\alpha\cdot \int_0^{\pi/3} \sec\beta \tan\beta \,d\beta \cdot \int_0^{\pi/3} \sec\gamma\tan\gamma\,d\gamma.
$$
One of these three will require integration by parts.  The other two are routine.
Postscript inspired by comments below:
Clearly
$$
\int_0^{\pi/3} \sec\beta\tan\beta\,d\beta = \sec\frac\pi3-\sec0 = 2-1=1.
$$
Then:
$$
\begin{align}
& \phantom{{}=}\int_0^{\pi/3}\alpha\Big(\sec\alpha\tan\alpha\,d\alpha\Big) \\[10pt]
& =\int\alpha\,d\delta=\alpha\delta-\int\delta\,d\alpha \\[10pt]
& =\alpha\sec\alpha-\int\sec\alpha\,d\alpha \\[10pt]
& =\left[\phantom{\int}\!\!\!\!\! \alpha\sec\alpha-\log \left| \sec\alpha + \tan\alpha \right| \, \right]_0^{\pi/3} \\[10pt]
& =\frac{2\pi}{3}-\log(2+\sqrt{3})
\end{align}
$$
Post-post-script inspired by further comments:
Remember that $\log(2+\sqrt{3})$ is the same thing as $-\log(2-\sqrt{3})$.
. . . and yet another addendum inspired by comments below:
Alright, so we've reduced the integral via trigonometric substitutions to
$$
\iiint \arcsin(\sin(\alpha+\beta+\gamma))  (\sec\alpha\tan\alpha\,d\alpha) (\sec\beta\tan\beta\,d\beta) (\sec\gamma\tan\gamma\,d\gamma)
$$
over the set $(\alpha,\beta,\gamma)\in[0,\pi/3]^3$.
If we can change $\arcsin(\sin(\bullet))$ to just $\bullet$, then the rest of what appears above solves the problem.  But it is objected that within $(\alpha,\beta,\gamma)\in[0,\pi/3]^3$ there are points where $\alpha+\beta+\gamma>\pi/2$, and so we'd have $\arcsin(\sin(\alpha+\beta+\gamma))= \pi-(\alpha+\beta+\gamma)$.
One can look at that, but before I do, I have to say I suspect it's quite unnatural to consider such a thing, so I wonder why it's being done.
To be continued . . .
