How to integrate $ \int e^{\cos( x)} \ dx$ I mainly did this for fun of it and am posting it here to have it reviewed and corrected if I made a mistake. This is a considerably simpler version of this solution I posted a couple months back.
Please comment with any corrections, questions, comments or concerns and hopefully you'll find it as interesting as I have! Also, I'm still fairly new to the community and I'm not certain I'm using the answer your own question format appropriately, so feel free to advise if that is the case.
Lastly, here's a link to the graphs on Desmos and a Mathematica function intended to be compared to NIntegrate!
eci[x_] := N[
Table[Sum[
Sum[Sin[x*(n - 2 k)]/((n - k)!k!(n - 2 k)*2^(n - 1)), {k, 0,
Floor[(n - 1)/2]}], {n, 1, L}] + BesselI[0, 1]*x, {L, 100, 100}]];
eciN[x_] := NIntegrate[E^Cos[z], {z, 0, x}];

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
Evaluating\ \int e^{cos( x)} \ dx:\\
Starting\ with\ the\ taylor\ series\ definition\ we\ have\ e^{cos( x)} =\ \sum _{n=0}^{\infty }\frac{cos^{n}( x)}{n!}\\
Now,\ using\ the\ complex\ exponential\ definition\ of\ cosine,\ we\ can\ obtain\ a\ generalized\ summation\\
for\ expanding\ integer\ powers\ of\ cos( x)\\
\\
cos( x) =\frac{e^{i\ x} +e^{-ix} \ }{2} \ \therefore \ cos^{n}( x) =\left(\frac{e^{i\ x} +e^{-i\ x} \ }{2}\right)^{n} =\frac{\left( e^{i\ x} +e^{-i\ x}\right)^{n}}{2^{n}}\\
\\
By\ evaluating\ the\ expansions\ we\ can\ obtain:\\
\\
\frac{\left( e^{i\ x} +e^{-i\ x}\right)^{n}}{2^{n}} =\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\binom{n}{k}\frac{cos( x\cdot ( n-2k))}{2^{n-1}} \ +\frac{cos^{2}\left(\frac{\pi n}{2}\right) \cdot \left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}}\\
\\
Now\ substituting\ the\ summation\ back\ into\ the\ taylor\ series,\ the\ function\ f( x) =\ e^{cos( x)} \ can\ now\\
be\ expressed\ by\ the\ following\ infinite\ summations:\\
\\
f( x) =\ e^{cos( x)} =1+\sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{cos( x( n-2k))}{2^{n-1} \cdot k!\cdot ( n-k) !} +\sum _{n=1}^{\infty }\frac{cos^{2}\left(\frac{\pi n}{2}\right) \cdot \left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}}\\
\\
Now\ separating\ the\ constant\ terms\ not\ dependent\ on\ x\ we\ have:\ 1+\sum _{n=1}^{\infty }\frac{cos^{2}\left(\frac{\pi n}{2}\right) \cdot \left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}}\\
Becuase\ the\ partial\ sums\ of\ \sum _{n=1}^{\infty }\frac{cos^{2}\left(\frac{\pi n}{2}\right) \cdot \left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}} \ are\ only\ non\ zero\ for\ even\ n,\\
we\ can\ rewrite\ this\ as\ 1+\ \ \sum _{n=1}^{\infty }\frac{1}{2^{2n} \cdot n!^{2}} =\sum _{n=0}^{\infty }\frac{1}{2^{2n} \cdot n!^{2}} =I_{0}( 1) ,\\
where\ I_{n}( z) \ is\ the\ modified\ Bessel\ function\ of\ the\ first\ kind.\\
\\
We\ can\ now\ express\ the\ function\ f( x) =e^{cos( x)} as:\ f( x) =\ I_{0}( 1) +\sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{cos( x( n-2k))}{2^{n-1} \cdot k!\cdot ( n-k) !}\\
\\
This\ now\ becomes\ the\ fairly\ simple\ integration:\\
F( x) =\int f( x) \ dx=\int I_{0}( 1) \ dx+\ \sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\int \frac{cos( x\cdot ( n-2k))}{( n-k) !\cdot k!\cdot 2^{n-1}} dx\\
\\
Finally,\ F( x) =\int e^{cos( x)} \ dx=x\cdot I_{0}( 1) \ +\ \sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{sin( x\cdot ( n-2k))}{( n-k) !\cdot k!\cdot ( n-2k) \cdot 2^{n-1}} +C\\
\\
Additionally,\ the\ sinusoid\ produced\ by\ the\ function\ g( x) =e^{cos( x)} -I_{0}( 1) \ can\ be\ expressed\ by:\\
g( x) =\sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{cos( x( n-2k))}{2^{n-1} \cdot k!\cdot ( n-k) !} \ and\ likewise;\\
G( x) =\int g( x) \ dx=\ \sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{sin( x\cdot ( n-2k))}{( n-k) !\cdot k!\cdot ( n-2k) \cdot 2^{n-1}} +C,\ shown\ in\ the\ figure\ below:\\
\end{array}$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
Finally,\ some\ other\ interesting\ take\ aways:\ \\
\\
-\ The\ maximum\ and\ minimum\ points\ of\ the\ sinusoid\ occur\ at:\ x=2\pi n\pm \left( \ 1+\frac{2}{\pi } -\frac{3}{\pi ^{2}}\right)\\
-\ F\left(\frac{\pi }{2}\right) =\ \int _{0}^{\frac{\pi }{2}} e^{cos( x)} \ dx\ =\frac{\pi ( I_{0}( 1) +L_{0}( 1))}{2} \ and\ G\left(\frac{\pi }{2}\right) =\int _{0}^{\frac{\pi }{2}}\left( e^{cos( x)} -I_{0}( 1)\right) \ dx=\ \frac{\pi \cdot L_{0}( 1)}{2}\\
\ \ \ \ ( \ Here,\ L_{n}( z) \ is\ the\ modified\ Struve\ function.)\\
\\
-\ Similarly\ to\ \int e^{cos( x)} \ dx,\ \int e^{sin( x)} \ dx\ can\ be\ evaulated\ by\ making\ the\ substitution\ x\rightarrow \left( x-\frac{\pi }{2}\right)\\
\ \ \ and\ adding\ \ \int _{0}^{\frac{\pi }{2}}\left( e^{cos( x)} -I_{0}( 1)\right) \ dx\ to\ obtain:\ \\
\ \ \ \int e^{sin( x)} \ dx=\ \frac{\pi \cdot L_{0}( 1)}{2} +x\cdot I_{0}( 1) +\ \sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{sin\left(\left( x-\frac{\pi }{2}\right) \cdot ( n-2k)\right)}{( n-k) !\cdot k!\cdot ( n-2k) \cdot 2^{n-1}} +C\\
\\
-\ A\ somewhat\ interesting\ take\ away\ may\ be\ that\ \\
\ \ \ L_{0}( 1) =\frac{2}{\pi }\sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{cos\left(\frac{\pi }{2} \cdot ( n-2k)\right)}{2^{n-1} \cdot k!\cdot ( n-k) !} =\sum _{n=0}^{\infty }\frac{\left(\frac{1}{2}\right)^{2n+1}}{\left(\left( n+\frac{1}{2}\right) !\right)^{2}}
\end{array}$
