<s>[INST] Double integral with coordinate transformation I want to integrate
$\int_B \cos\frac{x-y}{x+y}dB$
where the area B is characterized by the inequalities
$\frac{1}{2} - x \leq \frac{1}{2}$ 
$1 \leq 1 + y$.
$\frac{1}{2} \leq x + y \leq 1$
using the substitution $u = x - y, v = x + y$.
My attempt:
We have $x = \frac{u+v}{2}, y = \frac{v-u}{2}$.
So we get 
$\frac{1}{2} \leq v \leq 1$
$ 2-v \leq u \leq v$
which gives us
$\int_B \cos\frac{x-y}{x+y}dB = \int_{\frac{1}{2}}^1 \int_{2-v}^{v}   \cos\frac{u}{v}du \ dv = $
$\int_{\frac{1}{2}}^1 v \sin \frac{u}{v} \Big|_{2-v}^v dv = \int_{\frac{1}{2}}^1 v \sin 1 - v \sin\frac{2-v}{v} dv$
Somehow I feel I mixed something up with the domain of the inner integral. Is that true?
 [/INST] The region $B$ is bounded by the lines 
$$x=0, \quad y=0, \quad x+y = \frac{1}{2},\quad \textrm{and} \quad x+y = 1.$$
  Using your choice $x = \frac{u+v}{2}$ and $y = \frac{v-u}{2}$, these lines become 
$$u+v = 0,\quad  v-u = 0, \quad v = \frac{1}{2},\quad \textrm{and} \quad v=1.$$  Hence, we get $\frac{1}{2} \leq v \leq 1$ and $-v \leq u \leq v$.
</s>