Evaluating $\int e^{\sin x}\ dx$ in terms of a Taylor series. NOW before anyone gets too hasty in saying "oh, it's impossible," I would like to presume that it is possible despite the fact that it obviously is not, and theoretically search for some way of simply representing the solution. I came up with the idea of switching $e^{\sin x}$ into its taylor series and integrating every term. Now I have a new series, but how do I "de-Taylor" this series into some combination of elementary functions? Is this even possible with $e^{\sin x}$? Even if it isn't, is there a method for being able to de-Taylor a series? My equations:
$$
\int e^{\sin x}\ dx = x+\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{40}-\frac{x^6}{90}-\frac{x^7}{1680}\dots
$$
Surely there is some algorithm that satisfies those denominators, right? There just has to be! And of course, I know how naive this may seem to pursue, but I have taken an interest in this problem and am actively seeking a solution.
To get the power series for
$e^{\sin x}$,
you can do this:

Let 
$f(x) = e^{\sin x}$.
Then
$f'(x) 
= \cos x\,f(x)
$.

Therefore,
if
$f(x)
=\sum_{n=0}^{\infty}a_n x^n
$,
then
$f'(x)
=\sum_{n=1}^{\infty}n a_n x^{n-1}
=\sum_{n=0}^{\infty}(n+1) a_{n+1} x^{n}
$.

Since
$\cos(x)
=\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}
$,

$\begin{array}\\
f'(x)
&amp;=\cos(x)f(x)\\
&amp;=(\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!})(\sum_{n=0}^{\infty}a_n x^n)\\
&amp;=(\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!})(\sum_{n=0}^{\infty}a_n x^n)\\
&amp;=\sum_{m=0}^{\infty} x^m \sum_{k=0}^{\lfloor m/2 \rfloor} (-1)^k \frac{1}{(2k)!}a_{m-2k}\\
&amp;=\sum_{m=0}^{\infty} x^m \sum_{k=0}^{\lfloor m/2 \rfloor} (-1)^k \frac{a_{m-2k}}{(2k)!}\\
\end{array}
$

Therefore
$(n+1) a_{n+1}
=\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \frac{a_{n-2k}}{(2k)!}
$.

Since
$f(0) = 1$,
$a_0 = 1$.

Then,
with $n=0$,
$a_1
=1
$;
with $n=1$,
$2a_2
=a_1
=1
$
so
$a_2 = \frac12$;
with $n=2$,
$3a_3
=a_2-a_0/2
=\frac12-\frac12
=0
$;
with $n=3$,
$4a_4
=a_3-a_1/2
=0-\frac12
=-\frac12
$
so
$a_4 = -\frac18
$.

So 
$f(x)
=1+x+\frac{x^2}{2}-\frac{x^4}{8}+...
$.
Integrating,
$\int f(x)
=x+\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{40}+...
$.

This agrees with Wolfy
(there is an error in
your coefficient of $x^5$ -
you divided by 4 instead of 5).

Note that method
(which is not original by me)
can be used to get
the power series for
$e^{g(x)}$
where $g(0) = 0$:
let
$f(x) = e^{g(x)}$.
Then
$f'(x) 
= g'(x)e^{g(x)}
=g'(x)f(x)
$.
Substituting power series
for $f$ and $g$,
we get a recursion
for the coefficients
of $g$
in terms of the
coefficients of
$f$.

Read generatingfunctionology
(it's free!)
at
https://www.math.upenn.edu/~wilf/DownldGF.html
for lots more neat stuff.
There is no such thing as "de-Taylor". This integral can't be combined of elementary functions. Believe me because it's one of the simplest and most famous integrals that can't be combined of elementary functions. In fact only a small minority of integrals and taylor series have a presentation with elementary functions.