Volume of an ideal simplex in $\mathbb{H}^3$, idea/intuition behind result

A longer but more intuitive proof of this formula is by "integrating" the Schlaefli identity for ideal tetrahedra.

One starts from a euclidean triangle with angles $\alpha$, $\beta$, $\gamma$ and side lengths $a$, $b$, $c$. Put $x=\log a$, $y=\log b$, $z=\log c$ and consider the function $$f(x,y,z) = \alpha x + \beta y + \gamma z + \Lambda(\alpha) + \Lambda(\beta) + \Lambda(\gamma)$$ One computes (using that $x - \log(2\sin\alpha) = \log R$) the partial derivatives: $$\frac{\partial f}{\partial x} = \alpha, \quad \frac{\partial f}{\partial y} = \beta, \quad \frac{\partial f}{\partial z} = \gamma$$ Now take the Legendre transform of $f$: $$F(\alpha,\beta,\gamma) = \Lambda(\alpha)+\Lambda(\beta)+\Lambda(\gamma)$$ It follows that $$\frac{\partial F}{\partial \alpha} - \frac{\partial F}{\partial\beta} = y-x = \log\frac{b}{a} \quad \text{ etc.}$$ This $\log\frac{b}{a}$ turns out to be related to the truncated edge lengths of the tetrahedron, so that the function $F$ satisfies the Schlaefli differential identity. Therefore it is the volume up to an additive constant.

For more details see Sections 4 and 5 in

Bobenko, Alexander I.; Pinkall, Ulrich; Springborn, Boris A. Discrete conformal maps and ideal hyperbolic polyhedra. Geom. Topol. 19 (2015), no. 4, 2155–2215.

The arxiv version.

A variation of the other answer: first decompose an ideal tetrahedron into orthoschemes with two ideal vertices and (non-right) angles of the form $\alpha,\pi/2-\alpha, \alpha$.

These orthoschemes have volume a function of $\alpha$. To compute the variation, truncate the ideal vertices by horospheres meeting the other two vertices. Then as we vary $\alpha$, only one edge contributes to Schlafli's formula, and a simple geometric computation shows that the edge length is $\log(2\sin\alpha)$. Integrating gives the Lobachevsky function.

Note: The relevent version of Schlafli's formula was proved by Hodgson in his thesis. See the Remark on p. 125 of his thesis.