$\iint_D \cos \left( \frac{x-y}{x+y} \right)\,dA$ I'm studying to a Calculus test and I couldn't solve the following exercise:
Calculate $\iint_D \cos \left( \frac{x-y}{x+y} \right)\,dA$ where $D$ is the region of the plane $xy$ limited by $x+y=1$, $x=0$ and $y=0$.
After a little of thinking I've got the following:
$$
\int_0^1\int_0^{1-y}\cos\left(\frac{x-y}{x+y}\right)\,dx\,dy
$$
I have no idea how to solve it. Wolfram Alpha couldn't solve it too.
 A: Put $u=(x+y)/2$ and $v=(x-y)/2$. 
Reversing, $x=u+v,\ y=u-v$.
Then the border $x+y=1$ becomes $u=1/2$, while $x=0$ becomes $v=-u$ and $y=0$ becomes $v=u$. There will be a factor of 2 or 1/2 (I'll leave that to you to look up) because of the jacobian of the transformation, but the new setup of the iterated integral as
$$\int_0^{1/2} \int_{-u}^u cos(v/u)\ dv\ du$$ becomes doable and if i did it right comes out $\sin(1)/4$.
The topic to check is change of variables in a double integral, look for the "jacobian" of the transformation. Here the jacobian is definitely a constant, either 2 or 1/2, I just forgot which one to multiply the transformed integral by. However I think this is the way needed to get the integral.
ADDED: I just integrated your form numerically on maple, and comparing things found it agrees with the value $\sin(1)/2$. So looks like the integral jacobian factor multiplier is 2, rather than 1/2. This makes sense since the area of the new triangle iterated over is only half the original integrating triangle.
A: So, I've done the following:
$$
\iint_D \cos \left( \frac{x-y}{x+y} \right) \, dA, \,\,\,\, D: \{x+y=1, \, x=0, \, y=0\}
$$
Substitution:
$$
\alpha = x-y \therefore x = \frac{\alpha+\beta}{2} \\
\beta = x+y \therefore y = \frac{\beta-\alpha}{2} \\
J = \begin{vmatrix}\alpha_x & \alpha_y\\\beta_x & \beta_y\end{vmatrix}
= \begin{vmatrix}1 & -1 \\ 1 & 1\end{vmatrix} = 2
\\
$$
So, $x+y=1$ becomes $\beta = 1$, $x = 0$ becomes $\alpha = -\beta$ and $y = 0$ becomes $\alpha=\beta$. Let's setup the integral:
$$
\int_0^1 \int_{-\beta}^\beta \cos \left( \frac{\alpha}{\beta} \right) J \, d\alpha \, d\beta \\
= 2 \int_0^1 \int_{-\beta}^\beta \cos \left( \frac{\alpha}{\beta} \right) \, d\alpha \, d\beta
$$
Variable substitution:
$$
u = \frac{\alpha}{\beta} \\
du = \frac{d\alpha}{\beta}
$$
The integral becomes:
$$
2 \int_0^1 \int_{-1}^1 \cos u \, \beta \, du \, d\beta
$$
Solving:
$$
\begin{align*}
2 \int_0^1 \int_{-1}^1 \cos u \; \beta \; du \; d\beta
& = 2 \int_0^1 \left[ \sin u \, \beta \right]_{-1}^1 \, d\beta \\
&= 2 \int_0^1 2\beta\sin 1 \, d\beta \\
&= 2 \left[2 \sin 1 \frac{\beta^2}{2} \right]_0^1 \\
&= 2 \sin 1
\end{align*}
$$
