Cómo integrar ∫ecos(x) dx∫ecos⁡(x) dx \int e^{\cos( x)} \ dx

$\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}$

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