I've been puzzled by the following exercise:
Considering the set
$$D = \left\{ (x,y)\in \mathbb{R}^3: 1<x+y<2; 0<x<y \right\}$$
and the function $f:D\to \mathbb{R}$, $f(x,y) = (y^2-x^2)\cos(x+y)^4$. Calculate $\int_Df$, using an appropriate change of coordinates.
I've tried using polar coordinates to no avail (perhaps wrongly?). I'm guessing I have to use some other sort of coordinate system I'm unfamiliar with. Which one, then?
You can perform the change $u=x+y$ and $v=-x+y$. So, jacobian reads $dxdy=\frac{1}{2}dudv$ and new variables vary $1<u<2$ and $0<v<u$. Then, your integral
$$\int\int_D (y^2-x^2)\cos(x+y)^4dx\,dy=\frac{1}{2}\int_1^2\int_0^uv\,u\,\cos u^4\,dv\,du=\frac{1}{4}\int_1^2u^3 \,\cos u^4du=\frac{\sin 16-\sin 1}{16}$$