Answer
$x^2-\dfrac{x^6}{2!}+\dfrac{ x^{10}}{4!}-\dfrac{x^{14}}{6!}+..$
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^2 \cos (x^2)=x^2(1-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^4}{4!}-\dfrac{(x^2)^6}{6!}+...)=x^2-\dfrac{x^6}{2!}+\dfrac{ x^{10}}{4!}-\dfrac{x^{14}}{6!}+..$
