Worked solution: The Beukers–Calabi–Kolk double-integral proof (1993)

Step 4 of 6: The Jacobian cancels the denominator

In plain words

A change of variables always stretches area by a local scaling factor called the Jacobian; here that factor turns out, after a short trigonometric computation, to equal exactly the troublesome denominator $1-x^2y^2$ that started the whole problem, so it cancels it perfectly and leaves nothing behind.

\frac{\partial(x,y)}{\partial(u,v)}=1-\frac{\sin^2u\sin^2v}{\cos^2u\cos^2v}=1-x^2y^2

Detailed analysis

A direct computation of the four partial derivatives and the 2×2 determinant gives exactly $1-x^2y^2$ — the same expression sitting in the denominator of the integral. So $dx\,dy/(1-x^2y^2)=du\,dv$ , with no leftover factor at all.

Terms in this step

Jacobian determinant: The determinant $\partial(x,y)/\partial(u,v)$ of the matrix of partial derivatives of a change of variables; it measures how much a small patch of area is stretched or shrunk by the substitution.

Knowledge used in this step

Multiple integrals

Common mistake. It is tempting to stop at "the Jacobian is nonzero" and move on, but a valid change of variables also needs the map to be one-to-one on the open square — a separate fact (true here) that this computation alone does not establish.