An interesting solution of the Basel problem is given in https://www.tandfonline.com/doi/abs/10.1080/00029890.2018.1452473 (Basel Problem: A Solution Motivated by the Power of a Point, by Kapil R. Shenvi Pause).

In the integral $$\zeta(2)=\int_0^1\int_0^1\frac{dx\,dy}{1-xy}$$ let's make a change of variables $$x=\cos{\phi}-\tan{\theta}\sin{\phi},\;\;y=\cos{\phi}+\tan{\theta}\sin{\phi}. \tag{1}$$ The Jacobian of this transformation is twice the inverse of $1/(1-xy)$ and hence $$\zeta(2)=2\int\int d\theta\,d\phi.$$ The integration domain in the last integral is determined by the conditions $0\le x,y\le 1$, which gives $$-\frac{\phi}{2}\le\theta\le\frac{\phi}{2},\;\;\;\phi-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}-\phi. \tag{2}$$ Geometrically (2) is a quadrilateral in the $(\theta,\phi)$ plane with mutually orthogonal diagonals of lengths $\pi/3$ and $\pi/2$. Therefore $$\zeta(2)=2\;\frac{1}{2}\frac{\pi}{2}\frac{\pi}{3}=\frac{\pi^2}{6}.$$ Very nice, isn't it?

In the paper the transformation of variables (1) is motivated by an interpretation of $1/(1-xy)$ in terms of inversive geometry. Can this approach be extended to $$\zeta(n)=\int_0^1\cdots \int_0^1\frac{dx_1\cdots dx_n}{1-x_1\cdots x_n}?$$ In particular can we calculate in this manner $\zeta(2n)$? What is the relation (if any) of (1) with the Beukers-Kolk-Calabi change of variables? For the latter see https://pdfs.semanticscholar.org/35be/01e63c0bfd32b82c97d58ccc9c35471c3617.pdf

**P.S.** The Beukers-Kolk-Calabi change of variables is
$$x=\frac{\sin{\phi}}{\cos{\theta}},\;\;\;y=\frac{\sin{\theta}}{\cos{\phi}}.$$
Its Jacobian is $1-x^2y^2$ and it is applied to the integral
$$\zeta(2)=\frac{4}{3}\int_0^1\int_0^1\frac{dx\,dy}{1-x^2y^2}.$$
In this case the integration domain becomes isosceles right triangle
$0\le \phi,\;0\le \theta,\;\phi+\theta\le \pi/2$.