Compute $\iint_D(x^2-y^2)e^{2xy}dxdy$. Compute $$\iint_D(x^2-y^2)e^{2xy}dxdy,$$ where $D=\{(x,y):x^2+y^2\leq 1, \ -x\leq y\leq x, \ x\geq 0\}.$ 
The area is a circlesector disk with radius $1$ in the first and fourth quadrant. Going over to polar coordinates I get
$$\left\{
  \begin{array}{rcr}
    x & = & r\cos{\theta} \\
    y & = & r\sin{\theta} \\
  \end{array}, \ \ \implies E:\left\{
  \begin{array}{rcr}
    0 \leq r\leq 1 \\
    -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \\
  \end{array}
\right.
\right.$$
and $$J(r,\theta)=\frac{d(x,y)}{d(r,\theta)}=r.$$
So $$\iint_D(x^2-y^2)e^{2xy}r \ dxdy=\iint_Er^3(\cos^2{\theta}-\sin{2\theta})e^{r^22\cos{\theta}\sin{\theta}}drd\theta= \\ =\iint_Er^3\cos{2\theta} e^{r^2\sin{2\theta}}drd\theta = 2\int_0^{4/\pi}\cos{2\theta}\cdot\left(\int_0^1 r^3e^{r^2\sin{2\theta}}dr\right)d\theta.$$
I have no idea how to compute the inner integral. I seem to get quite complicated integrals everytime I do this.

By letting $x=\frac{u+v}{\sqrt{2}},y=\frac{u-v}{\sqrt{2}}$ the given integral becomes
$$\begin{eqnarray*} \iint_{\substack{u^2+v^2\leq 1 \\ u,v\geq 0}} 2uv\,e^{u^2-v^2}\,du\,dv &=& \int_{0}^{1}\int_{0}^{\pi/2}\rho^3\sin(2\theta)e^{\rho^2\cos(2\theta)}\,d\theta\,d\rho\\&=&\frac{1}{2}\int_{0}^{1}\int_{0}^{\pi}\rho^3\sin(\theta)e^{\rho^2\cos(\theta)}\,d\theta\,d\rho\\&=&\frac{1}{2}\int_{0}^{1}2\rho \sinh(\rho^2)\,d\rho\\&=&\frac{1}{2}\left[\cosh(\rho^2)\right]_{0}^{1}=\frac{\cosh(1)-1}{2}=\color{red}{\sinh^2\left(\tfrac{1}{2}\right)}.\end{eqnarray*}$$
