Conjectured integral for Catalan's constant

There is also an interpretation of this integral in terms of hyperbolic geometry. In hyperbolic geometry, Catalan's constant $G$ is one quarter the (three dimensional) volume of a regular ideal octahedron (or the volume of an ideal tetrahedron with dihedral angles $\frac{\pi}{2}, \frac{\pi}{4}$, and $\frac{\pi}{4}$).

If $\Gamma$ is the group of orientation preserving isometries of a tessellation of $\mathbb{H}^3$ by regular ideal octahedra (aka $PGL(2,O_1)$), the quotient of $\mathbb{H}^3/\Gamma$ has volume: $$\frac{G}{6}=\int_0 ^\frac{1}{2}\int_0 ^\frac{1}{2}\int_\sqrt{1-x^2-y^2} ^\infty \frac{1}{z^3} dz dy dx = \frac{1}{2}\int_0 ^\frac{1}{2}\int_0 ^\frac{1}{2}\frac{1}{1-x^2-y^2} dy dx .$$

A decent reference for understanding this observation is Neumann and Reid's Notes on Adams' small volume orbifolds (page 312). Although not directly stated, their method relies on the observation that a certain manifold, the Whitehead link complement, is well known to be isometric to a regular ideal octahedron with faces identified in pairs, as noted above this means it has volume $4G$. The orbifold we are interested in is a 24 fold quotient of this manifold, and so it has volume $\frac{G}{6}$. The geometry of the orbifold is described in Colin Adams' paper Noncompact 3-Orbifolds of Small Volume (see figure 6(c) and Theorem 5.2).

However, an early reference is Borel's paper: Commensurability classes and volumes of hyperbolic 3-manifolds.

Mathematica confirms the following:

Change the integral to polar coordinates to get

$$\frac{(1)}{2} = \int_0^{\pi/4} \int_0^{\sec(\theta)/2} \frac{1}{1-r^2}r\ dr\ d\theta = \frac{G}{6}.$$

After some trick substitutions, I've put the integral in the form:

$$\int_0^{\pi/6}{\coth^{-1}{(2\,\cos{t})}dt}$$

Mathematica (version 8) then returns the exact value $\frac{G}{3}$. A nice definite integral for Catalan's constant, anyway.