Evaluate
$$\iint_R (y-x)\sin((y+x)^3)dydx$$
where $R$ is the triangular region with vertices $(0, 0), (2, 2), (0, 4)$, by making a suitable change of variables.
What I have so far is that $u=y-x$ and $v=y+x$. From there I got $y=(u+v)/2$ and $x=(u-v)/2$.
Edit: I got the double integral as $\int_{0}^{4}\int_{v}^{0} u\sin(v^3)|-\frac{1}{2}|dudv$ = $\frac{1}{12}\left(-\cos \left(64\right)+1\right)$
Let $u=\frac1{\sqrt2}(x+y)$ and $v=\frac1{\sqrt2}(y-x)$, which rotates the coordinates 45-degrees. The vertexes $(2,2)$ becomes $(2\sqrt2,0)$ and $(0,4)$ becomes $(2\sqrt2,2\sqrt2)$. As a result, the integral is,
$$\iint_R (y-x)\sin((y+x)^3)dydx$$ $$=\int_0^{2\sqrt2}\int_0^u \sqrt2 v\>\sin(2^{3/2}u^3) dvdu$$ $$=\frac1{2^{1/2}}\int_0^{2\sqrt2}u^2\sin(2^{3/2}u^3) du$$ $$=\frac1{12}\int_0^{2\sqrt2}d\left(-\cos(2^{3/2}u^3) \right)$$ $$=-\frac1{12}\cos(2^{3/2}u^3)\bigg|_0^{2\sqrt2}$$ $$= \frac1{12}(1-\cos 64)=\frac16\sin^2 32$$