Evaluating $ \int_0^\theta \cosh(a\sin x) dx$ 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 $0 < \theta < \pi/2$.
 A: 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).
A: $\int_0^\theta\cosh(a\sin x)~dx=\int_0^\theta\sum\limits_{n=0}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}~dx=\int_0^\theta\left(1+\sum\limits_{n=1}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}\right)~dx$
For $n$ is any natural number,
$\int\sin^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\therefore\int_0^\theta\left(1+\sum\limits_{n=1}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}\right)~dx$
$=\left[\sum\limits_{n=0}^\infty\dfrac{a^{2n}x}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{a^{2n}((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}\right]_0^\theta$
$=\sum\limits_{n=0}^\infty\dfrac{a^{2n}\theta}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{a^{2n}((k-1)!)^2\sin^{2k-1}\theta\cos\theta}{4^{n-k+1}(n!)^2(2k-1)!}$
$=\theta I_0(a)-\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{a^{2n}((k-1)!)^2\sin^{2k-1}\theta\cos\theta}{4^{n-k+1}(n!)^2(2k-1)!}$
$=\theta I_0(a)-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n+2k+2}(k!)^2\sin^{2k+1}\theta\cos\theta}{4^{n+1}((n+k+1)!)^2(2k+1)!}$
Or you can express in terms of Incomplete Bessel Functions:
Consider $$\begin{align}J_0(ia,w)&=\dfrac{2}{\pi}\int_0^w\cos(ia\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^w\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cosh\left(a\cos\left(x-\dfrac{\pi}{2}\right)\right)~d\left(x-\dfrac{\pi}{2}\right)
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cosh(a\sin x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_\frac{\pi}{2}^0\cosh\left(a\sin\left(\dfrac{\pi}{2}-x\right)\right)~d\left(\dfrac{\pi}{2}-x\right)
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_\pi^\frac{\pi}{2}\cosh(a\cos(\pi-x))~d(\pi-x)
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_\frac{\pi}{2}^\pi\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\pi\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-I_0(a)\end{align}
$$
Then $\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx=I_0(a)+I_0(a,w)$
$\therefore\int_0^\theta\cosh(a\sin x)~dx=\dfrac{\pi}{2}\left(I_0(a)+I_0\left(a,\theta-\dfrac{\pi}{2}\right)\right)$
A: As we see in the other answers, there is no closed-form for this indefinite integral.
However there is a closed form for certain definite integrals.  For example
$$
\int_0^{\pi/2} \cosh(a\sin \theta)\;d\theta = \frac{\pi}{2}\;I_0(a)
$$
where $I_0$ is the hyperbolic Bessel function of order $0$

Similar answers:
https://math.stackexchange.com/a/2829346/442
https://math.stackexchange.com/a/1844794/442 
