Answer
$x+x^2+\dfrac{1}{3}x^3-\dfrac{1}{30} x^5-...$
Work Step by Step
We know that the Taylor Series for $e^x$ and $\sin x$ is defined as:
$e^x =1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+....$ and $\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+..$
Here, we have
$e^x \sin x=(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+....)(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+..)$
or, $=x+x^2+\dfrac{1}{3}x^3-\dfrac{1}{30} x^5-...$
