Triangle integral with vertices 
Evaluate $$I=\iint\limits_R \sin \left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\, dA,$$ where $R$ is the triangle with vertices $(0,0),(2,0)$ and $(1,1)$.
Hint: use $u=\dfrac{x+y}{2},v=\dfrac{x-y}{2}$.

Can anyone help me with this question I am very lost. Please help
I know you can make the intergal $\sin(u)\cos(v)$, but then what to do?

$\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=\frac{1}{2}\left(\sin x+\sin y\right)$
The line joining $(0,0)$ and $(2,0)$ has an equation $y=0$ and $0\leq x\leq 2$
The second line: $y=-x+2$
The third line: $y=x$
The integral becomes:
$$I=\frac{1}{2}\left(\int\limits_{0}^{1}\int\limits_{0}^{x}+\int\limits_{1}^{2}\int\limits_{0}^{2-x}\right)(\sin x+\sin y)dy~dx=\frac{1}{2}\left(\int\limits_{0}^{1}\int\limits_{0}^{x}(\sin x+\sin y)dy~dx+\int\limits_{1}^{2}\int\limits_{0}^{2-x}(\sin x+\sin y)dy~dx\right)=\frac{1}{2}\left(\int\limits_0^1\left[y\sin x-\cos y\right]_0^xdx+\int\limits_1^2\left[y\sin x-\cos y\right]_0^{2-x}dx\right)=\frac{1}{2}\left(\left[\sin x-x\cos x+x-\cos x\right]_0^1+\left[(x-2)\cos x-\sin x+x+\sin(2-x)\right]_1^2\right)$$ 
