integrate this double integral by any method you can. I'm having trouble with this double integral:
$$\int_0^2\int_0^{2-x} \exp\left(\frac{x−y}{x+y}\right)\text dy\,\text dx$$
 A: The integral is one over the $2$-simplex $\Delta_2(2) = \{ (x,y ) : x, y \ge 0, x+ y \le 2 \}$.  
One standard trick to deal with integral over $d$-simplex of the form
$$\Delta_d(L) = \{ (x_1, x_2, \ldots, x_d ) : x_i \ge 0, \sum_{i=1}^d x_i  \le L \}$$
is convert it to one over the $d$-cuboid $[0,L] \times [0,1]^{d-1}$ through following change of variables
$$\begin{align}
\lambda &= x_1 + x_2 + \cdots + x_d\\
\lambda\mu_1 &= x_1 + x_2 + \cdots + x_{d-1}\\
\lambda\mu_1\mu_2 &= x_1 + x_2 + \cdots + x_{d-2}\\
&\;\;\vdots\\
\lambda\mu_1\mu_2\cdots\mu_{d-1} &= x_1
\end{align}
$$
Under such change of variables, an integral of $(x_1,\ldots,x_d)$ over $\Delta_d(L)$ becomes an integral of $(\lambda,\mu_1,\ldots,\mu_{d-1})$ over 
$[0,L] \times [0,1]^{d-1}$.
For the integral at hand, let $\lambda = x + y$ and $x = \mu\lambda$.
The area element can be rewritten as
$$dx \wedge dy = dx \wedge d(x+y) = d(\mu \lambda) \wedge d\lambda = \lambda d\mu \wedge d\lambda$$
So the integral becomes
$$\int_{\Delta_2(2)} \exp\left(\frac{x-y}{x+y}\right)dxdy
= \int_0^2 \int_0^1 e^{2\mu - 1} \lambda d\mu d\lambda
= \left(\int_0^2 \lambda d\lambda \right)\left(\int_0^1 e^{2\mu-1} d\mu\right)\\
= 2 \times \frac{1}{2e}(e^2 - 1)
= 2\sinh(1)
\approx 2.35040238728760291376476370119120163
$$
