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?

Following the hint we obtain a new coordinate system rotated $\frac{\pi}{4}$ clockwise.
See the following picture.

A unit area in the new system is the double of the old one.
Thus the integral changes to:
$$2 \int_0^{1} \int_v^1{\sin u \cos v \, \mathrm {d}u \, \mathrm {d}v}$$
which you can easily finish.
