Sum of Reciprocals of Squares of Odd Integers/Proof 4

Proof
Applying the substitution:
 * $\tuple {x, y} = \tuple {\dfrac {\map \sin u} {\map \cos v}, \dfrac {\map \sin v} {\map \cos u} }$

the Jacobian matrix is:
 * $\mathbf J_{\mathbf f} := \begin{pmatrix}

{\dfrac \partial {\partial u} \dfrac {\map \sin u} {\map \cos v} } & {\dfrac \partial {\partial v} \dfrac {\map \sin u} {\map \cos v} } \\ {\dfrac \partial {\partial u} \dfrac {\map \sin v} {\map \cos u} } & {\dfrac \partial {\partial v} \dfrac {\map \sin v} {\map \cos u} } \end{pmatrix}$

and the Jacobian determinant is:

Under this substitution, the image of the region $\closedint 0 1^2$, that is the unit square, is an isosceles triangle, $\bigtriangleup$ with base and height $\dfrac \pi 2$: Vertices: $\tuple {0, 0}; \tuple {0, \dfrac \pi 2}; \tuple {\dfrac \pi 2, 0}$.

We have:


 * $ 0 \le \dfrac {\map \sin u} {\map \cos v} \le 1$
 * $\leadsto u \ge 0$ and
 * $\leadsto \map \sin u \le \map \cos v$

And we have:


 * $ 0 \le \dfrac {\map \sin v} {\map \cos u} \le 1$
 * $\leadsto v \ge 0$ and
 * $\leadsto \map \sin v \le \map \cos u$

This gives us the region: $\set {\tuple {u, v}: u, v \ge 0 \land \map \cos u \ge \map \sin v \land \map \cos v \ge \map \sin u}$.

Which is equivalent to the region: $\set {\tuple {u, v}: u, v \ge 0 \land v \le \dfrac \pi 2 - u }$.

Hence,: