sunshine022uv

2022-01-03

Find out
$\int {e}^{\mathrm{cos}x}dx$

Robert Pina

First of all: there is no close form solution in terms of elementary functions.
What can you do, but its

Anzante2m

This is just a response to your comments about having tried integration by parts for this problem.
Youre

karton

$\int {e}^{\mathrm{cos}x}dx\phantom{\rule{0ex}{0ex}}={\int }_{n=0}^{\mathrm{\infty }}\frac{{\mathrm{cos}}^{2n}x}{\left(2n\right)!}dx+\int \sum _{n=0}^{\mathrm{\infty }}\frac{{\mathrm{cos}}^{2n+1}x}{\left(2n+1\right)!}dx\phantom{\rule{0ex}{0ex}}=\int \left(1+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{cos}}^{2n}x}{\left(2n\right)!}\right)dx+\int \sum _{n=0}^{\int }\frac{{\mathrm{cos}}^{2n+1}x}{\left(2n+1\right)!}dx\phantom{\rule{0ex}{0ex}}\text{For n is any natural number,}\phantom{\rule{0ex}{0ex}}\int {\mathrm{cos}}^{2n}xdx=\frac{\left(2n\right)!x}{{4}^{n}\left(n!{\right)}^{2}}+\sum _{k=1}^{n}\frac{\left(2n\right)!\left(\left(k-1\right)!{\right)}^{2}\mathrm{sin}x{\mathrm{cos}}^{2k-1}x}{{4}^{n-k+1}\left(n!{\right)}^{2}\left(2k-1\right)!}+C\phantom{\rule{0ex}{0ex}}\text{This result can be done by successive integration by parts.}\phantom{\rule{0ex}{0ex}}\text{For n is any non-negative integer,}\phantom{\rule{0ex}{0ex}}\int {\mathrm{cos}}^{2n+1}xdx\phantom{\rule{0ex}{0ex}}=\int {\mathrm{cos}}^{2n}xd\left(\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}=\int \left(1-{\mathrm{sin}}^{2}x{\right)}^{n}d\left(\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}=\int \sum _{k=0}^{\mathrm{\infty }}{C}_{k}^{n}\left(-1{\right)}^{k}{\mathrm{sin}}^{2k}xd\left(\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}=\sum _{k=0}^{n}\frac{\left(-1{\right)}^{k}n!{\mathrm{sin}}^{2k+1}x}{k!\left(n-k\right)!\left(2k+1\right)}+C\phantom{\rule{0ex}{0ex}}\therefore \int \left(1+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{cos}}^{2n}x}{\left(2n\right)!}\right)dx+\int \sum _{n=0}^{\mathrm{\infty }}\frac{{\mathrm{cos}}^{2n+1}x}{\left(2n+1\right)!}dx\phantom{\rule{0ex}{0ex}}=x+\sum _{n=1}^{\mathrm{\infty }}\frac{x}{{4}^{n}\left(n!{\right)}^{2}}+\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^{n}\frac{\left(k-1\right)!{\right)}^{2}\mathrm{sin}x{\mathrm{cos}}^{2k-1}x}{{4}^{n-k+1}\left(n!{\right)}^{2}\left(2k-1\right)!}+\sum _{k=0}^{n}\frac{\left(-1{\right)}^{k}n!{\mathrm{sin}}^{2k+1}x}{\left(2n+1\right)!k!\left(n-k\right)!\left(2k+1\right)}+C$