Why is $\frac{1}{4}\int_{-1}^{1}\int_{-1}^{1}e^{it(x+y)}(1+xy(x^{2}-y^{2})) dxdy =\frac{\sin^{2}(t)}{t^{2}}$ 
Why is
  $\displaystyle{{1 \over 4}\int_{-1}^{1}\int_{-1}^{1}
\mathrm{e}^{\mathrm{i}t\left(x + y\right)}\
\left[1 + xy\left(x^{2} - y^{2}\right)\right]\,\mathrm{d}x\,\mathrm{d}y
=
{\sin^{2}\left(t\right) \over t^{2}}}$ ?.

How do I solve this integral ?. The answer must be right and WolframAlpha gives the same solution.
But calculating
$\displaystyle{\int_{-1}^{1}\mathrm{e}^{\mathrm{i}t\left(x + y\right)}
\left[1 + xy\left(x^{2} - y^{2}\right)\right]\mathrm{d}x}$ first will probably give my a really complicated term, so I guess there must be some kind of tricky subsitution or identity that I can't see right now.

The square $[-1,1]^2$ is symmetric with respect to its diagonals, hence the given integral equals the same integral with the variables $x$ and $y$ exchanged. By cancellation it follows that
$$ \iint_{(-1,1)^2}e^{it(x+y)}(1+xy(x^2-y^2))\,dx\,dy = \iint_{(-1,1)^2}e^{it(x+y)}\cdot 1\,dx\,dy $$
and by Fubini's theorem the last integral is the square of $\int_{-1}^{1}e^{itz}\,dz$, i.e. $4\frac{\sin^2 t}{t}$.
