Integration by parts seems to be infinite, but answer is simple

I have to integrate this:

$$\int_{-1}^0 \int_{-2x-2}^{2x+2} x^2 y^2 + \sin(xy) e^{x^2} y^2 \;\operatorname d\!y \operatorname d\!x$$

I started integrating by parts,

$$\int_{-1}^0 e^{x^2} \left({-y^2 \cos(xy)\over x} - \int_{-2x-2}^{2x+2} {-2y \cos(xy)\over x} \operatorname d\!y \right) \operatorname d\!x$$

but it seems to go on and on and on. It stops after two, but I have to do the same for $x$. Also, I have this other integral:

$$\int_0^1 \int_{2x-2}^{-2x+2} x^2 y^2 + \sin(xy) e^{x^2} y^2 \;\operatorname d\!y \operatorname d\!x$$

I did it with Maple. It took 1-2 seconds and I got $4 \over 45$. It usually takes way longer even if it's a bit complicated.

Is there any trick I could use? Did I miss anything?

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You may have already been aware of this, but be careful when using integration by parts with definite integrals. There shouldn't be a $y$ outside of your $\mathrm dy$ integral on your second line. – Mark S. Feb 16 '14 at 1:31
Can you please explain? $u \operatorname d\!v = u v - \int v \operatorname d\!u$, right? – ChrisVolkoff Feb 16 '14 at 2:07
That fact tells you about indefinite integrals. But you have to remember to plug in the limits of integration to the $uv$ part when you have a definite integral. In other words, $\int_a^bu\mathrm dv=uv\mid_a^b-\int_a^bv\mathrm du$. In particular, your $-y^2\cos(xy)/x$ term should have something like $\mid_{y=-2x-2}^{y=2x+2}$ next to it so you don't forget to plug in. (Separately, a variable outside of an integral with respect to that variable should be a red flag; at best it means you're using a variable in two different ways, but it usually means you've made an error.) – Mark S. Feb 16 '14 at 2:21
I totally forgot about evaluating $uv$! Thanks for pointing it out, I'll try to remember. – ChrisVolkoff Feb 16 '14 at 2:39

The "difficult" part of your integrand, namely $\sin(xy) e^{x^{2}} y^{2}$, is an odd function integrated over a symmetric interval.
Oh my, of course! $$\int_{-2x-2}^{2x+2} \sin(xy) e^{x^2} y^2 \;\operatorname d\!y = 0$$ Thank you very much, this simplifies everything! – ChrisVolkoff Feb 16 '14 at 0:09