Answer
$x^3-\dfrac{1}{3!}x^5+\dfrac{1}{5!}x^7-\dfrac{1}{7!}x^9+...$
Work Step by Step
Consider the Maclaurin Series for $\sin x$ is defined as:
$ \sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Multiply the above series with $x^2$ as follows:
Then $x^2 \sin x=x^2 (x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)$
$\implies x^3-\dfrac{1}{3!}x^5+\dfrac{1}{5!}x^7-\dfrac{1}{7!}x^9+...$
