Calculating $\int_{0}^{\pi}\text{sinh}\left(\sin\left(x\right)\right)\text{d}x$ I was wondering if it exists a beautiful exact value to
$$ \int_{0}^{\pi}\text{sinh}\left(\sin\left(x\right)\right)\text{d}x$$
which has a nice graph over $\left[0,\pi\right]$, but i can't get to compute it.
 A: If you know the Beta and Gamma functions well, you can see the sequence $$a_n:=\int_0^\pi\sin^{2n+1}x\mathrm{d}x=2\int_0^{\pi/2}\sin^{2n+1}x\mathrm{d}x=\operatorname{B}\left(\frac{1}{2},\,n+1\right)=\frac{n!\sqrt{\pi}}{\Gamma\left(n+\frac{3}{2}\right)}$$satisfies$$a_0=2,\,\frac{a_{n+1}}{a_n}=\frac{2(n+1)}{2n+3}$$so by induction$$a_n=2^{n+1}\prod_{i=1}^{n-1}\frac{i}{2i-1}=\frac{2^{2n+1}n!^2}{(2n+1)!}.$$So your integral is $$2\sum_{n\ge 0}\frac{n!^2}{(2n+1)!^2}4^n=\pi L_0(1),$$with $L_n(x)$ the modified Struve function. (I'm not an expert on this function myself, but Wolfram Alpha gives the result, and I suspect it can be derived from the previous link's Eq. (1) using Legendre's duplication formula.)
A: Using the integral representation of the modified Struve function linked by J.G. which is given by

$$\mathbf L_\nu(z)~=~\frac{2\left(\frac12 z\right)^\nu}{\sqrt \pi~\Gamma\left(\nu+\frac12\right)}\int_0^{\pi/2}\sinh(z\cos t)\sin^{2\nu}t~\mathrm dt\tag1$$

By setting $z=1,\nu=0$ and after reflection on $\frac\pi2$ we almost obtain the desired integral. Precisely we got that
\begin{align*}
\mathbf L_0(1)&=\frac{2\left(\frac12 1\right)^0}{\sqrt \pi~\Gamma\left(0+\frac12\right)}\int_0^{\pi/2}\sinh(1\cdot\cos t)\sin^{2\cdot 0}t~\mathrm dt\\
\mathbf L_0(1)&=\frac1{ \pi}2\underbrace{\int_0^{\pi/2}\sinh(\cos t)~\mathrm dt}_{t\mapsto \frac\pi2-t}\\
\mathbf L_0(1)&=\frac1{ \pi}2\int_0^{\pi/2}\sinh(\sin t)~\mathrm dt
\end{align*}
On the other hand the original integral is symmetric towards $x=\frac\pi2$. Thus
$$\int_0^\pi\sinh(\sin t)~\mathrm dt=\int_0^{\pi/2}\sinh(\sin t)~\mathrm dt+\underbrace{\int_{\pi/2}^\pi\sinh(\sin t)~\mathrm dt}_{t\mapsto \pi-t}=2\int_0^{\pi/2}\sinh(\sin t)\mathrm dt$$

$$\therefore~\int_0^\pi\sinh(\sin t)~\mathrm dt~=~2\int_0^{\pi/2}\sinh(\sin t)\mathrm dt~=~\pi\mathbf L_0(1)$$

I guess closer to an actual closed-form we will not get. However, on the other hand the modified Struve function has got a series represenation given by

$$\mathbf L_\nu(z)~=~\left(\frac12 z\right)^{\nu+1}\sum_{n=0}^\infty\frac{\left(\frac12 z\right)^{2n}}{\Gamma\left(n+\frac32\right)\Gamma\left(n+\nu+\frac32\right)}\tag2$$

Again setting $z=1$ and $\nu=0$ to get 
\begin{align*}
\mathbf L_0(1)&=\left(\frac12 1\right)^{0+1}\sum_{n=0}^\infty\frac{\left(\frac12 1\right)^{2n}}{\Gamma\left(n+\frac32\right)\Gamma\left(n+0+\frac32\right)}\\
\mathbf L_0(1)&=\frac12\sum_{n=0}^\infty\frac{4^{-n}}{\Gamma^2\left(n+\frac32\right)}\\
\mathbf L_0(1)&=\frac12\sum_{n=0}^\infty\frac{4^{-n}}{\left(n+\frac12\right)^2\Gamma^2\left(n+\frac12\right)}\frac{\Gamma^2\left(\frac12\right)}{\Gamma^2\left(\frac12\right)}\\
\mathbf L_0(1)&=\frac2\pi\sum_{n=0}^\infty\frac{4^{-n}}{\left(2n+1\right)^2}\left(\frac{\Gamma\left(\frac12\right)}{\Gamma\left(n+\frac12\right)}\right)^2\\
\mathbf L_0(1)&=\frac2\pi\sum_{n=0}^\infty\frac{4^{-n}}{\left(2n+1\right)^2}\left(\frac{4^nn!}{(2n)!}\right)^2\\
\mathbf L_0(1)&=\frac2\pi\sum_{n=0}^\infty\frac{4^n (n!)^2}{(2n+1)^2(2n)!^2}
\end{align*}

$$\therefore~2\sum_{n=0}^\infty\frac{4^n (n!)^2}{(2n+1)!^2}~=~\pi\mathbf L_0(1)$$

The latter agrees perfectly wiht J.G. sum representation aswell as with the integral representation.
A: The integrand is not integratable in terms of elementary function but we have the following series representation:
$$\int_{0}^{\pi}\text{sinh}\left(\sin\left(x\right)\right)\text{d}x=2\cdot\sum_{k=0}^\infty \Bigg(\frac{2^k \cdot k!}{(2k+1)!}\Bigg)^2$$
This comes from using the series representation of $\sinh{(\sin{(x)})}$ which is the following:
$$\sinh{(\sin{(x)})}=\sum_{k=0}^\infty\frac{\sin^{2k+1}(x)}{(2k+1)!}$$
and integrating both sides with respect to $x$ in the given range.
