Transform the region and compute the integral The solution for the integral $I=\int_{0}^{1}\int_{0}^{1}\frac{1}{1-x^2 y^2}dxdy$ is given by transformation
$$
x=\frac{\sin u}{\cos v}\qquad \text{and}\qquad y=\frac{\sin v}{\cos u}
$$
and then integral becomes  $I=\int\int_{E}dudv$, where $E$ is the trinagle with vertices $(0,0)$, $(\frac{\pi}{2},0)$ and $(0,\frac{\pi}{2})$.
I can not see how the square $\{(x,y):0\le x,y\le 1\}$ transform to the triangle $E$ under this transform?
Since $x,y\geq 0$, both sine and cosine should have same sign. It should also be $\sin u\le \cos v$ and $\sin v\le \cos u$ since $x,y\leq 1$.
For example let $A=\{(0,y):0\le y\le 1\}$ be one side of square. Under this transformation $0=x=\frac{\sin u}{\cos v}\rightarrow \sin u=0\rightarrow \cos u=1 \text{  or } \cos u=-1$. In first case,  $\cos u=1\rightarrow y=\sin v\rightarrow 0\le \sin v\le 1\rightarrow v\in [0,\frac{\pi}{2}]+2k\pi$. In second case, $\cos u=-1\rightarrow y=-\sin v\rightarrow -1\le \sin v\le 0\rightarrow v\in [-\frac{\pi}{2},0]+2k\pi$. I can not see the line segment $A$ transform which shape under this transformation?
 A: $$(u,v) \leftrightarrow (x,y)  \ \ \text{with} \ x=\frac{\sin u}{\cos v} \ \text{and} \ \ y=\frac{\sin v}{\cos u} \tag{1}$$
Here is a graphical answer obtained with Geogebra:

To the "sweeping" of the triangle by red line segments $R_m$ with equations
$$x+y=m \ \ \text{for} \ \ 0 < m < \pi/2$$
is associated the "sweeping" of the square $[0,1]^2$ by black curves $B_m$ with parametric equations
$$x(t)=\frac{\sin(t)}{\cos(m-t)}, \ \ y(t)=\frac{\sin(m-t)}{\cos(t)}$$
At bottom left, the first red lines are almost identical to their black images (indeed for small angles, formulas (1) give $x\approx\frac{u}{1}$ and $y\approx\frac{v}{1}$). Further on, the black curves, images of the red line segments, display an  increasing bending.
In fact two other simple sweepings are possible as illustrated in the figure below

$$y-x=n  \ \ \text{for} \ \ -\pi/2 < n < \pi/2$$
in correspondence with the curves whose parametric equations are:
$$x(t)=\frac{\sin(t)}{\cos(n+t)}, \ \ y(t)=\frac{\sin(n+t)}{\cos(t)}$$
Besides, I advise you to see the very interesting exchange here ; in particular, your change of variables is given in the answer by Vivek Kaushik.
