Integral of cosine over a triangle I need to integrate $\pi \cos (\pi x)$ over a triangle T with vertices $(0,0)$, $(1,1)$ and $(2,0)$.
Me reasoning is:
$$\pi \int_T \cos (\pi x) dxdy =\pi\int_0^1 \int_0^{y=x}\cos (\pi x) dxdy+\pi\int_1^2 \int_{y=2-x}^0\cos (\pi x) dxdy=...=2\sin \pi=0$$
The only non-immediate step is $\pi\int x \cos (\pi x) = x\sin x+\frac{1}{\pi}\cos (\pi x)$.
Are the integral expression or the result correct? Both? Neither?
Thanks in advance!

HINT:
Save a step by integrating over $x$ first as
$$\pi\int_0^1\left(\int_{y}^{2-y}\cos (\pi x)\,dx\right)\,dy=\int_0^1\left(\left.\sin(\pi x)\right|_{y}^{2-y}\right)\,dy$$
Can you finish from here?
