Just to add to my previous answer the actual formula so one does not have to follow the site
$$\displaystyle \int e^{\sin(x)} dx=I_0(1)x + \frac{\pi}{2}L_0(1) + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nx - \frac{n\pi}{2} \right ) $$
Another nice way of solving definite integral apart for simply stating its value through Struve and Bessel (which is the shortest possible known expression at the moment) goes like this:
First let us get rid of ##\sin(x)##, introducing ##u=\sin(x), du=\cos(x)dx## This leads to
$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=2\int_{0}^{1} \frac{e^u}{\sqrt{1-u^2}} du$$
Notice that we have taken it twice from ##0## to ##\frac{\pi}{2}## as ##e^{\sin(x)}## is symmetrical.
Now we use expansion of ##e^u## reducing it all to the sum of integrals
$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=2 \sum_{k=0}^{\infty} \int_{0}^{1} \frac{u^k}{k!\sqrt{1-u^2}} du$$
Now
$$\displaystyle \int_{0}^{1} \frac{u^k}{k!\sqrt{1-u^2}} du= \frac{1}{k!}\frac{\sqrt{\pi}\Gamma(\frac{k+1}{2})}{2\Gamma(\frac{k}{2}+1)}$$
coming from the connection between Beta and Gamma function, making it all
$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=\sum_{k=0}^{\infty} \frac{\sqrt{\pi}\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2}+1)k!}$$
or in a split form
$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=\sum_{k=0}^{\infty}\frac{\pi}{4^n(n!)^2} + \sum_{k=0}^{\infty} \frac{2^{n+1}n!}{(2n+1)!(2n+1)!}$$
First 10 terms are giving 20 digit precision already.
I am happy with 4 terms 4 digit precision
$$\displaystyle \frac{328}{147} + \frac{2917 π}{2304} \approx 6.2087$$
Just to make the connection
$$\displaystyle \pi L_0(1)=\sum_{k=0}^{\infty} \frac{2^{n+1}n!}{(2n+1)!(2n+1)!}$$
$$\displaystyle \pi I_0(1)=\sum_{k=0}^{\infty}\frac{\pi}{4^n(n!)^2}$$
