The integral below seems quite simple, but I couldn't find anywhere the result. $$ I = \int_0^\theta \cosh(a\sin x) dx$$ I tried to expand it into Taylor expansion series and successfully evaluate the integral, but it just got mess, $$ I =\sum_{k=0}^{\infty} \frac{a^{2k}}{(2k)!} \left[ \frac{1}{2^{2k}}\binom{2k}{k}\theta + \frac{(-1)^k}{2^{2k-1}}\sum_{n=0}^{k-1}(-1)^n\binom{2k}{n} \frac{\sin[(2k-2n)\theta]}{2k-2n}\right]. $$
Is there any simpler form of this integral? Any helps or hints will be appreciated!
Edited: $\theta$ can only have value of $0 < \theta < \pi/2$.
This antiderivative is not an elementary function. However, at $\theta = \pi$ the integral is $\pi I_0(a)$ where $I_0$ is a modified Bessel function.
EDIT: With the substitution $x = \arcsin(t)$, the integral becomes $$ I = \int_0^{\sin(\theta)} \dfrac{\cosh(at)\; dt}{\sqrt{1-t^2}} = \sum_{k=0}^\infty {2k \choose k} 4^{-k} \int_0^{\sin(\theta)} t^{2k} \cosh(at)\; dt $$ Let $\sin(\theta) = s$. Now $$ \eqalign{\int_0^s t^{2k} \cosh(at)\; dt &= \dfrac{d^{2k}}{da^{2k}} \int_0^s \cosh(at)\; dt =\dfrac{d^{2k}}{da^{2k}} \dfrac{\sinh(as)}{a}\cr &= \dfrac{1}{2a^{2k+1}} \left(\Gamma(2k+1,-as) - \Gamma(2k+1,as)\right)}$$ so $$ I = \sum_{k=0}^\infty {2k \choose k} \dfrac{\Gamma(2k+1,-as) - \Gamma(2k+1,as)}{(2a)^{2k+1}} $$ but I don't know a closed form for that sum (nor does Maple).