Solution
= Solution
Use the <Maclaurin series> $\sin x=x-x^3/6+O(x^5)$ and $e^u=1+u+u^2/2+u^3/6+u^4/24+O(u^5)$. The terms through fourth degree give
$$
\begin{aligned}
e^{\sin x}
&=1+\left(x-\frac{x^3}{6}\right)
+\frac12\left(x^2-\frac{x^4}{3}\right)
+\frac{x^3}{6}+\frac{x^4}{24}+O(x^5)\\
&=\boxed{1+x+\frac{x^2}{2}-\frac{x^4}{8}+O(x^5).}
\end{aligned}
$$
In particular the cubic coefficient cancels; it has not been omitted accidentally.