How to integrate $\cos \left(\frac{y-x}{y+x}\right)$? Suppose $D$ is a trapezoid with vertices $(0, 1)$, $(1, 0)$, $(0, 2)$, $(2, 0)$. How to calculate the following?
$$\iint_D \cos \left(\frac{y-x}{y+x}\right) \, \mathrm d x \mathrm d y$$
P.S.
Thanks a lot to Robert Z, I solved the problem now. My solution is this:
$$u=y-x$$
$$v=y+x$$
$$\iint_Dcos(\frac{y-x}{y+x})dxdy = \int_1^2 \int_{-v}^v \cos(\frac{u}{v}) du dv = \int_1^2 2v\sin(1) dv = 3\sin(1)$$

Hint. Let $u=y−x$ and $v=x+y$. Then
$$\iint_D \cos \left(\frac{y-x}{y+x}\right) \, d x d y=\iint_T \cos \left(\frac{u}{v}\right) \, J(u,v)d u d v.$$
What is the new domain of integration $T$. What is $J(u,v)$? Can you take it from here?
