Answer
$x^3-\dfrac{x^5}{3!}+\dfrac{x^7}{5!}-\dfrac{x^9}{7!}+...$
Work Step by Step
Since, we know that the Maclaurin Series for $\sin x$ is defined as:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Now, $x^2 \sin x=x^2 (x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)=x^3-\dfrac{x^5}{3!}+\dfrac{x^7}{5!}-\dfrac{x^9}{7!}+...$
