Answer
$=x^3-\dfrac{x^5}{3}+\dfrac{2 x^7}{45}+....$
Work Step by Step
The Maclaurin series for $\sin x$ can be defined as:
$x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......=\Sigma_{n=0}^{\infty} \dfrac{ (-1)^n x^{2n+1}}{(2n+1)!}$
Now, $x \times \sin^2 x=x [\Sigma_{n=0}^{\infty}\dfrac{ (-1)^n
x^{2n+1}}{(2n+1)!}]^2$
$\implies x \times \sin^2 x=x[x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......]^2$
Thus, we have the result:
$x \times \sin^2 x =x^3-\dfrac{x^5}{3}+\dfrac{2 x^7}{45}+....$