$\int_0^1 \int_0^{1-y} \cos\Big( \frac{x-y}{x+y} \Big) \, dx dy$

Question

Reviewing old homework sets for a class and I came across this integral:

$$\displaystyle \int_0^1 \int_0^{1-y} \cos\Big( \frac{x-y}{x+y} \Big) \; dx dy,$$

which the question suggests to evaluate using a change of coordinates; however, I haven't a clue where to begin to identify a useful change of coordinates.

I tried $u = x-y$ and $v = x+y$, but then wasn't sure how I'd convert the domain of integration.

After that, I looked to the given limits for inspiration and noticed that $0<x<1-y$ could be rewritten $y < x+y < 1$, so tried $u = x+y$ and $v = y$, which yielded

$$\displaystyle \int_0^1 \int_y^{1} \cos\Big( \frac{ u - 2y }{u} \Big) \; du\,dy,$$

but that doesn't seem any simpler than the original, to me.

Any advice would be appreciated!

Answer 1

Your original substitution was right. The original domain of integration is the triangle bounded by $x=0$, $y=0$, and $x+y=1$.

Since $x = \frac12(u+v)$, $y=\frac12(v-u)$, and $x+y=v$, these boundary curves correspond to the curves $v=-u$, $v=u$, and $v=1$ in the $uv$-plane.

Can you finish now? (You need to stop to consider whether you prefer to integrate $du\,dv$ or $dv\,du$.)

Answer 2

Note that by $u = x-y$ and $v = x+y$

$$du\,dv=|J| \,dx \,dy \iff du\,dv=\begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}dx\,dy\iff dx\,dy=\frac12du\,dv$$

then

$$\int_0^1 \int_0^{1-y} \cos\Big( \frac{x-y}{x+y} \Big) \; dx dy=\frac12\int_0^1 dv \int_{-v}^{v} \cos\Big( \frac{u}{v} \Big) \; du$$