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?
 A: There is another way to do this.  You might notice that the integrand is
$$2 \sin{\left(\frac{x+y}{2}\right)} \cos{\left(\frac{x-y}{2}\right)} = \sin{x} + \sin{y}$$
You may then integrate this over the triangle directly:
$$\frac12 \int_0^1 dx \, \int_0^x dy \, [\sin{x} + \sin{y}] + \frac12 \int_1^2 dx \, \int_0^{2-x} dy \, [\sin{x} + \sin{y}] $$
Note that I formed the integration boundaries from the equations of the lines formed from the vertices of the triangle.  Note also that I had to break this in two: one for the left side and one for the right.
You may then evaluate this in terms of single integrals by integrating over $y$; I get
$$\frac12\int_0^1 dx \, [x \, \sin{x} + 1 - \cos{x}] + \frac12\int_1^2 dx \, [(2-x) \sin{x} + 1-\cos{(2-x)}]$$
I will let you take it from here.
A: Here is how you advance
1) Solve the system
$$u=\dfrac{x+y}{2},\,v=\dfrac{x-y}{2} $$
for $x$ and $y$.
2) Find the Jacobian  
$$ J=\begin{bmatrix} \dfrac{\partial x}{\partial u} \quad \dfrac{\partial x}{\partial v} \\  & \\ \dfrac{\partial y}{\partial u} \quad  \dfrac{\partial y}{\partial v} \end{bmatrix}. $$ 
3) $$ dxdy = |J|dudv .$$
4) Find the limits of integration for $u$ and $v$.
A: $\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=\frac{1}{2}\left(\sin x+\sin y\right)$
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:
$$I=\frac{1}{2}\left(\int\limits_{0}^{1}\int\limits_{0}^{x}+\int\limits_{1}^{2}\int\limits_{0}^{2-x}\right)(\sin x+\sin y)dy~dx=\frac{1}{2}\left(\int\limits_{0}^{1}\int\limits_{0}^{x}(\sin x+\sin y)dy~dx+\int\limits_{1}^{2}\int\limits_{0}^{2-x}(\sin x+\sin y)dy~dx\right)=\frac{1}{2}\left(\int\limits_0^1\left[y\sin x-\cos y\right]_0^xdx+\int\limits_1^2\left[y\sin x-\cos y\right]_0^{2-x}dx\right)=\frac{1}{2}\left(\left[\sin x-x\cos x+x-\cos x\right]_0^1+\left[(x-2)\cos x-\sin x+x+\sin(2-x)\right]_1^2\right)$$ 
A: 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.
