Answer
$x-\dfrac{\pi^2}{2!}x^3+\dfrac{\pi^4}{4!} x^5-\dfrac{\pi^6}{6!} x^7+..$
Work Step by Step
Consider the Maclaurin Series for $\cos x$ is defined as:
$ \cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$
Then, $ x \cos \pi x=x(1-\dfrac{(\pi x)^2}{2!}+\dfrac{(\pi x)^4}{4!}-\dfrac{(\pi x)^6}{6!}+...)$
$\implies x-\dfrac{\pi^2}{2!}x^3+\dfrac{\pi^4}{4!} x^5-\dfrac{\pi^6}{6!} x^7+..$
