Answer
$x^3-\dfrac{x^5}{3}+\dfrac{2 x^7}{45}+....$
Work Step by Step
The Maclaurin's series for $\sin x$ is as follows:
$x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......=\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}$
Therefore, $f(x)=x \sin^2 x$
$x \sin^2 x=x [\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}]^2$
or, $=x[x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......]^2$
or, $=x^3-\dfrac{x^5}{3}+\dfrac{2 x^7}{45}+....$
