$\int_0^\theta\cosh(a\sin x)~dx=\int_0^\theta\sum\limits_{n=0}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}~dx=\int_0^\theta\left(1+\sum\limits_{n=1}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}\right)~dx$
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 $\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx=I_0(a)+I_0(a,w)$
$\therefore\int_0^\theta\cosh(a\sin x)~dx=\dfrac{\pi}{2}\left(I_0(a)+I_0\left(a,\theta-\dfrac{\pi}{2}\right)\right)$