Answer
$1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+....$
Work Step by Step
Recall the Taylor series for $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
$ e^{\sin x} =1+\sin x+\dfrac{(\sin x)^2}{2!}+.....$
or, $ e^{\sin x}=1+(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-....) +\dfrac{1}{2 !} (x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-....) +....$
or, $ e^{\sin x} =1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+....$
