Answer
$5-\dfrac{5 \pi^2x^2}{2!}+\dfrac{5 \pi^4x^4}{4!}-\dfrac{5 \pi^6x^6}{6!}+..$
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)!}$
Now,
$5 \cos (\pi x)=5 \Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi x)^{2n}}{(2n)!}$
or, $ =5-\dfrac{5 \pi^2x^2}{2!}+\dfrac{5 \pi^4x^4}{4!}-\dfrac{5 \pi^6x^6}{6!}+..$
