# Trigonometric substitution in double integral

1. Mar 29, 2013

### drawar

1. The problem statement, all variables and given/known data

Let $R = \{ (x,y) \in \mathbb{R^{2}}: 0<x<1, 0<y<1\}$ be the unit square on the xy-plane. Use the change of variables $x = \frac{{\sin u}}{{\cos v}}$ and $y = \frac{{\sin v}}{{\cos u}}$ to evaluate the integral $\iint_R {\frac{1} {{1 - {{(xy)}^2}}}dxdy}$

2. Relevant equations

3. The attempt at a solution

I've already computed the Jacobian:

$\frac{{\partial (x,y)}} {{\partial (u,v)}} = \left| {\begin{array}{*{20}{c}} {\frac{{\partial x}} {{\partial u}}} & {\frac{{\partial x}} {{\partial v}}} \\ {\frac{{\partial y}} {{\partial u}}} & {\frac{{\partial y}} {{\partial v}}} \\ \end{array} } \right| = \left| {\begin{array}{*{20}{c}} {\frac{{\cos u}} {{\cos v}}} & {\frac{{\sin u\sin v}} {{{{\cos }^2}v}}} \\ {\frac{{\sin u\sin v}} {{{{\cos }^2}u}}} & {\frac{{\cos v}} {{\cos u}}} \\ \end{array} } \right| = 1 - {\left( {\frac{{\sin u\sin v}} {{\cos u\cos v}}} \right)^2}$

Now I'm left with finding out how $R$ would look like in the uv-plane. Hope someone can shed some light on this, thanks!

2. Mar 30, 2013

### SammyS

Staff Emeritus
I presume that the image of R, the unit square in x & y, is a subset of the set $\displaystyle \ \left\{ (u,\,v)\left|\, 0<u<\frac{\pi}{2}\,,\ 0<v<\frac{\pi}{2}\right.\right\}\ .$

If x = 0, then u = 0 and v ≠ π/2 .

If x = 1, then u = sin-1(cos(v)) = sin-1(sin(π/2 - v)) = π/2 - v .

etc.

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