How to integrate $\cos \left(\frac{y-x}{y+x}\right)$?

Question

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: $$u=y-x$$ $$v=y+x$$ $$\iint_Dcos(\frac{y-x}{y+x})dxdy = \int_1^2 \int_{-v}^v \cos(\frac{u}{v}) du dv = \int_1^2 2v\sin(1) dv = 3\sin(1)$$

Answer 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?

Answer 2

I think I would have preferred listing the points of the vertices of the trapezoid in the order $(0,1), (1,0), (2,0), (0,2)$

Sketch the regoin you'll see this

Note the top is $x+y=2$ and the bottom is $x+y=1$

So you can evaluate the integral by one of the following

$$\int_1^2\int_{1-x}^{2-x} \cos \left(\frac{y-x}{y+x}\right) dydx$$

$$\int_1^2\int_{1-y}^{2-y} \cos \left(\frac{y-x}{y+x}\right) dxdy$$

Unfortunately neither of those is particularly easy to evaluate the antiderivative. So consider the transformation $u=y-x$ and $v=y+x$

This transforms the points to $(1,1),(−1,1),(−2,2),(2,2)$ and the region to what we'll call $R$ below

The Jacobian $J(u,v)= \begin{bmatrix} \partial u/ \partial x & \partial u/ \partial y \\ \partial v/ \partial x & \partial v/ \partial y \\ \end{bmatrix} =\begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} $

Has $|J(u,v)|=-2$

So your integral becomes $$\iint_R \cos\left(\frac{u}{v}\right) (-2) dA$$

It looks a little messy since you'll need 3 different parts if you want to do it in rectangular coordinates. Polar coordinates might give some fun options but I haven't worked them through.