Answer
The co-efficient of $x^3$ is $0$and the co-efficient of $x^4$ is $\dfrac{e}{6}$.
Work Step by Step
Write the Taylor series for the function as: $\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+......$
Replace $x$ by $t^2$ in the above series to obtain:
$e^{\cos x}=e^{(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+......)}=e[e^{(-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+......)}]$
Therefore, we find that the co-efficient of $x^3$ is $0$and the co-efficient of $x^4$ is $e(\dfrac{1}{4!}+\dfrac{1}{2^3})=\dfrac{e}{6}$.
