Surface Integral over a rhombus Evaluate the integral

$$\int\int_{R}(x-y)^2 cos^2(x+y)dxdy$$ where $R$ is the rhombus with successive vertices as $(\pi,0), (2\pi,\pi), (\pi,2\pi), (0,\pi).$

My attempt- I tried doing this surface integral by performing integration by parts but the answer solution seems to be very lengthy, also am unable to get the final answer by this method. Just wanted to know if there is any other simpler method or can it be solved by using some theorem in vector calculus perhaps Green's theorem or something.

Thanks
1) Translate rhombus to origin:
$$
x' = x - \pi \quad y' = y - \pi
$$
$$
\int\int_{R}(x-y)^2 cos^2(x+y)dxdy \rightarrow \int\int_{R'}(x'-y')^2 cos^2(x'+y' + 2\pi)dx'dy'
$$
2) Change coordinate system:
$$
x' - y' = u \quad x' + y' = v
$$
Now:
$$
-\pi \leq u \leq \pi \quad -\pi \leq v \leq \pi 
$$
$$
\int\int_{R'}(x'-y')^2 cos^2(x'+y')dx'dy' \rightarrow \int\int_{R''}u^2 cos^2(v)J(u,v)dudv
$$
3) Compute Jacobian: $J = 1/2$

4) Evaluate Integral:
$$
\frac{1}{2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}u^2 cos^2(v)dudv = \frac{1}{2}\int_{-\pi}^{\pi} u^2 du \int_{-\pi}^{\pi}cos^2(v)dv = \frac{1}{2}\frac{2\pi^3}{3} \cdot \pi =  \frac{\pi^4}{3}
$$
This isn’t a surface integral because you’re integrating over a planar region R. Find a transformation that maps the rhombus in the xy-plane to a square in the uv-plane, and compute it’s jacobian. The problem incentivizes you to use the change of coordinates u = x-y, v = x+y (changes in coordinates like these are valuable because they tend to preserve the linearity of the transformation, which makes the problem easier to compute). The rhombus is a closed curve so Green’s Theorem applies, though I’d apply the change of coordinates first and then use the theorem. Keep in mind that doing so would force you to evaluate four separate line integrals.