Integration help - question: $e^{-\sin(x)}$ I would really like some help with the integration of $e^{-\sin(x)}$. Thanks to anyone who will help :)
Given that $\sin(x) > \frac{2x}{\pi}$ for $0 < x < \frac{\pi}{2}$, where $$\int_0^{\pi/2}e^{-\sin x}\,dx<\int_0^{\pi/2}e^{-2x/\pi}\,dx$$
RTS: 
$$\int_0^{\pi/2}e^{-\sin x}\,dx=\int_{\pi/2}^{\pi}e^{-\sin x}\,dx$$
 A: We have:
$$I=\int_{0}^{\pi/2}e^{-\sin x}\,dx = \int_{0}^{1}\frac{e^{-t}}{\sqrt{1-t^2}}\,dt$$
and since:
$$e^{-t}=\sum_{k\geq 0}\frac{(-1)^k t^k}{k!},\qquad \int_{0}^{1}\frac{t^k}{\sqrt{1-t^2}}\,dt =\int_{0}^{\pi/2}\sin^k\theta\,d\theta=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{2\,\Gamma\left(\frac{k}{2}+1\right)}$$
$($see Wallis' integrals for more information$)$, it follows that:
$$ I = \sum_{k\geq 0}\frac{(-1)^k \Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{2\,\Gamma\left(\frac{k}{2}+1\right)\Gamma(k+1)}=\frac{\pi}{2}\sum_{k\geq 0}\frac{(-1)^k}{2^k \Gamma\left(\frac{k}{2}+1\right)^2}=\color{red}{\frac{\pi}{2}\left(I_0(1)-L_0(1)\right)},$$
where $I_0$ and $L_0$ are a Bessel and a Struve function.

As to why the integrand does not possess an anti-derivative expressible in terms of elementary functions, see Liouville's theorem and the Risch algorithm.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{}$
\begin{align}&\overbrace{\color{#66f}{\int_{0}^{\pi/2}\expo{-\sin\pars{x}}\,\dd x}}
^{\ds{\dsc{\sin\pars{x}}=\dsc{t}\ \imp\ \dsc{x}=\dsc{\arcsin\pars{t}}}}\ =\
\int_{0}^{1}\frac{\expo{-t}}{\root{1 - t^{2}}}\,\dd t
=\color{#66f}{\large -\,\frac{\pi}{2}\,{\rm M}_{0}\pars{1}}
\approx{\tt 0.8731}\tag{1}
\end{align}
$\ds{\,{\rm M}_{\nu}\pars{z}}$ is a
Modified Struve Function.
The result
$\ds{\pars{~\mbox{given in expression}\ \pars{1}~}}$ corresponds to $11.5.4$ in
this link.
