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.

Call $\alpha = \sin(2\theta)$ for simplicity.
$$\int_0^1 r^3 e^{\alpha r^2}\ dr = \int_0^1 \frac{d}{d\alpha} r e^{\alpha r^2} = \frac{d}{d\alpha} \int_0^1 r e^{\alpha r^2}\ dr$$
The latter can easily be done by parts once to get
$$\int r e^{\alpha r}\ dr = \frac{e^{a r} (a r-1)}{a^2}$$
Hence with the extrema it becomes
$$\frac{e^a (a-1)+1}{a^2}$$
Hence
$$\frac{d}{d\alpha} \left(\frac{e^a (a-1)+1}{a^2}\right) = \frac{\left(a^2-2 a+2\right) e^a-2}{a^3}$$
Getting back $\alpha = \sin(2\theta)$ and thou hast
$$\frac{\left(\sin^2(2\theta)-2 \sin(2\theta)+2\right) e^{\sin(2\theta)}-2}{\sin^3(2\theta)}$$
