Consider the double integral over a region $D$ $$\iint_D \cos\left(\frac{x-y}{x+y}\right) \mathrm{dA}$$ where $D$ is the triangular region in the $xy$-plane with vertices $\{(0, 0), (0, 1), (1, 0)\}$. Sketch the image of the region $D$ in the $uv$-plane under the variable transformation $u = x - y, v = x + y$ and evaluate the double integral using the new variables $u$ and $v$.
I found the new region $S$ to be a triangular region which represented an isosceles triangle with vertices at $(0,0), (0,2) \text{ and } (1,1)$. The variable transformation also meant that $ x = \frac{1}{2}(v+u)$ and $y = \frac{1}{2}(v-u)$. I know that $$\int\int f(x,y) \text{dydx}$$ over a region $R$ also equals to $$\int\int f(x(u,v), y(u,v))\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \\\end{vmatrix}\text{dvdu}$$ over a region $S$.
By substituting in all the equations, I got: $$\int _0 ^1 \int _u ^{-u+2} \frac{1}{2}\cos\left(\frac{u}v{}\right)\text{dvdu}$$ I changed the order of integration and got: $$\int _0 ^2 \int _v ^{2-v} \frac{1}{2}\cos\left(\frac{u}v{}\right)\text{dudv}$$ which equals: $$\frac{1}{2} \times \int _0 ^2 v(\sin \left(\frac{2-v}{v}\right) - \sin (1))\text{dv}$$ which I don't know how to simplify any further but I'm sure there is a rational answer. Can anyone help me with this problem?
Your bounds of integration are incorrect. Using the transformation, we have $(0,0)\to (0,0)$, $(0,1)\to (-1,1)$, $(1,0)\to (1,1)$. This is the region $-1\le u\le 1,|u|\le v\le 1$; since your substitution is linear, it is determined by three points. The Jacobian determinant is correct, as is the new integrand. So we have $$ I=\int_{-1}^{1} \int_{|u|}^1\cos\left(\frac{u}{v}\right) \cdot \frac{1}{2}\;dvdu $$Now change the order of integration and integrate: $$ I=\frac{1}{2}\int_{0}^{1} \int_{-v}^v\cos\left(\frac{u}{v}\right) \;dudv =\int_{0}^{1} v\sin\left(1\right) \;dv=\frac{\sin(1)}{2} $$