Answer
$\frac{1}{3 \pi}$
Work Step by Step
$\int_{0}^{1 / 2} \int_{0}^{\pi} x \cos (x y) \cos ^{2} \pi x d y d x=\iint_{R} x \cos (x y) \cos ^{2} \pi x d A$
$$
\begin{aligned}=& \int_{0}^{1 / 2}\left[\sin (x y) \cos ^{2} \pi x\right]_{0}^{\pi} d x \\ &=\int_{0}^{1 / 2} \sin (\pi x) \cos ^{2} \pi x d x \end{aligned}
$$
Substitute $u=\cos \pi x$ and $d u=-\pi \sin \pi x d x$
$=-\frac{1}{\pi} \int_{1}^{0} u^{2} d u$
$=-\frac{1}{\pi}\left[\frac{u^{3}}{3}\right]_{1}^{0}=\frac{1}{3 \pi}$
