Answer
$\dfrac{1}{4!}x^4-\dfrac{1}{6!}x^6+\dfrac{1}{8!}x^8-\dfrac{1}{10!}x^{10}+...$
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
$\dfrac{x^2}{2}-1+\cos x=\dfrac{x^2}{2}-1+1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$
or, $\implies \dfrac{1}{4!}x^4-\dfrac{1}{6!}x^6+\dfrac{1}{8!}x^8-\dfrac{1}{10!}x^{10}+...$
