Answer
$$1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+....$$
Work Step by Step
The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
We have
$ e^{\sin x} =1+\sin x+\dfrac{(\sin x)^2}{2!}+.....$
or, $=1+(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-....) +\dfrac{1}{2 !} (x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-....) +....$
or, $=1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+....$
