Answer
$1-\dfrac{5^2}{2!}x^4+\dfrac{5^4}{4!}x^8-\dfrac{5^6}{6!}x^{12}...$
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, $ \cos(5x^2)=1-\dfrac{(5x^2)^2}{2!}+\dfrac{(5x^2)^4}{4!}-\dfrac{(5x^2)^6}{6!}+...$
or, $=1-\dfrac{5^2}{2!}x^4+\dfrac{5^4}{4!}x^8-\dfrac{5^6}{6!}x^{12}...$
