Answer
$$\frac{1}{2\pi} - \frac{1}{\pi^2}$$
Work Step by Step
Let $u$=x, $dv+\cos(\pi x) \ dx$. Then $du=1$ and $v=\frac{1}{\pi}\sin (\pi x)$.
Apply integration by parts to find the indefinite integral.
$$\int x\cos (\pi x) \ dx= \frac{x}{\pi} \sin(\pi x)- \int \frac{1}{\pi} \sin(\pi x) \ dx$$
$$=\frac{x}{\pi} \sin(\pi x) +\frac{1}{\pi ^2} \cos (\pi x) +C$$
Evaluate for $x=0, \frac {1}{2}$
$$=\frac{x}{\pi} \sin(\pi x) +\frac{1}{\pi ^2} \cos (\pi x) \Big]^\frac{1}{2}_0 =\frac{1}{2\pi} - \frac{1}{\pi^2}$$
