Answer
$x-\dfrac{\pi^2 x^3}{2!}+\dfrac{\pi^4 x^5}{4!}-\dfrac{\pi^6 x^7}{6!}+..$
Work Step by Step
Since, we know that the Maclaurin Series for $\cos x$ is defined as:
$ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$
Now,
$ x \cos \pi x=x(1-\dfrac{(\pi x)^2}{2!}+\dfrac{(\pi x)^4}{4!}-\dfrac{(\pi x)^6}{6!}+...)=x-\dfrac{\pi^2 x^3}{2!}+\dfrac{\pi^4 x^5}{4!}-\dfrac{\pi^6 x^7}{6!}+..$
