Change of variables in double integral. Finding limits of integration 
I need to integrate $\int_ \! \int \sin \frac{1}{2}(x+y) \cos\frac{1}{2}(x-y)\,dx\,dy$ over region $R$:{triangle with vertices $(0,0),(0,2),(1,1)$}. They ask to use $u=\frac{1}{2}(x+y)$ and $v=\frac{1}{2}(x-y)$.

Attempt:First, I transformed $(x,y)$ to $(x=x(u,v),y=y(u,v))$. Namely, I solved for x and y:
$$\begin{cases}u=\frac{1}{2}(x+y)\\v=\frac{1}{2}(x-y)\end{cases}$$
The Jacobian I found is $J(u,v)=\frac{\partial (x,y)}{\partial (u,v)}=-1$.
I am having hard time founding the limits of integration. In xy-plane $R$ looks like that:

So, the region R is bounded by $\begin{cases} y=0\\y=x\\y=-x+2 \end{cases}$
In uv-plane it looks like:

The region S is bounded by $\begin{cases} u=1\\ u=v\end{cases}$
Now the double integral looks like:
$$\int_0^1 \! \int_0^v \sin u \cos v\,du\,dv$$
When, I solve it I get 
$$\int_0^1 \! \int_0^v \sin u \cos v\,du\,dv=\frac{1}{2} (\frac{1}{2} \sin2 -1)$$
But in the answer key the answer is $1-\frac{1}{2} \sin2 $
Can you please tell me what I am doing wrong. Hints please.

Two issues: (1) $|J| = 2$, as mentioned above. (2) Your $S$ region should be flipped  about the $u$ axis. Check by computing the $u$-$v$ coordinates of $(0,0), (0,2), (1,1)$.
