What is the integral of $e^{\cos x}$ Question:

Find out $\displaystyle{\int e^{\cos x}~dx}$.

My Attempt:
Let $\cos x = y$. Hence $-\sin x\ dx = dy$ or $$dx = \displaystyle{\frac{-dy}{\sin x}=\frac{-dy}{\sqrt{1-\cos^2x}}=\frac{-dy}{\sqrt{1-y^2}}}$$ So
$$\begin{align}\int e^{\cos x}~dx &= \int e^y\left(\frac{-dy}{\sqrt{1-y^2}}\right)\\
&=-\int\frac{e^y}{\sqrt{1-y^2}}~dy
\end{align}$$
This integral is one I can't solve. I have been trying to do it for the last two days, but can't get success. I can't do it by parts because the new integral thus formed will be even more difficult to solve. I can't find out any substitution that I can make in this integral to make it simpler. Please help me solve it. Is the problem with my first substitution $y=\cos x$ or is there any other way to solve the integral $\displaystyle{\int\frac{e^y}{\sqrt{1-y^2}}~dy}$?

$\int e^{\cos x}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n}x}{(2n)!}dx+\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n+1}x}{(2n+1)!}dx$
$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\cos^{2n}x}{(2n)!}\right)dx+\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n+1}x}{(2n+1)!}dx$
For $n$ is any natural number,
$\int\cos^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
For $n$ is any non-negative integer,
$\int\cos^{2n+1}x~dx$
$=\int\cos^{2n}x~d(\sin x)$
$=\int(1-\sin^2x)^n~d(\sin x)$
$=\int\sum\limits_{k=0}^nC_k^n(-1)^k\sin^{2k}x~d(\sin x)$
$=\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}x}{k!(n-k)!(2k+1)}+C$
$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\cos^{2n}x}{(2n)!}\right)dx+\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n+1}x}{(2n+1)!}dx$
$=x+\sum\limits_{n=1}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
