Answer
$\Sigma_{n=0}^\infty (-1)^n\dfrac{2^{2n}x^{2n+1}}{(2n+1)!}$
Work Step by Step
We know that $\sin x \cos x=\dfrac{\sin 2x}{2}$
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!}+...$
Then $\sin x \cos x=\dfrac{\sin 2x}{2}=\dfrac{2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-\dfrac{(2x)^7}{7!}+...}{2}$
or, $=x-\dfrac{4x^3}{3!}+\dfrac{16x^5}{5!}-\dfrac{64x^7}{7!}+...$
or, $=\Sigma_{n=0}^\infty (-1)^n\dfrac{2^{2n}x^{2n+1}}{(2n+1)!}$
