The following is just a trial and it is not guaranteed correct.
(Please don't trust my calculation too much as I am very tired and may commit careless mistake.)
Note that we can write $a\sin\phi+b\cos\phi=r\sin(\phi+\phi_{0})$,
where $r=\sqrt{a^{2}+b^{2}}$, and $\phi_{0}\in[0,2\pi)$.
The actual value of $\phi_{0}$ is not important. Note that the integrand
is periodic with period $2\pi$. Therefore
\begin{eqnarray*}
& & \int_{0}^{2\pi}\exp(-C)\exp(a\sin\phi+b\cos\phi)d\phi\\
& = & \int_{\phi_{0}}^{2\pi+\phi_{0}}\exp(-C)\exp(r\sin\phi)d\phi\\
& = & \int_{0}^{2\pi}\exp(-C)\exp(r\sin\phi)d\phi.
\end{eqnarray*}
The last equality holds because $\int_{\phi_{0}}^{2\pi+\phi_{0}}=\int_{0}^{2\pi}+\int_{2\pi}^{2\pi+\phi_{0}}-\int_{0}^{\phi_{0}}=\int_{0}^{2\pi}$, with the second and the third integrals cancel each other.
Note that $\exp(x)=1+x+\frac{x^{2}}{2!}+\cdots$ and the series converges
uniformly on $[-r,r]$. Therefore, we can integrate termwisely and
get
\begin{eqnarray*}
& & \int_{0}^{2\pi}\exp(r\sin\phi)d\phi\\
& = & \sum_{k=0}^{\infty}\frac{r^{k}}{k!}\int_{0}^{2\pi}\sin^{k}\phi d\phi.
\end{eqnarray*}
If $k$ is odd, $\int_{0}^{2\pi}\sin^{k}\phi d\phi=0$ because $\int_{\pi}^{2\pi}\sin^{k}\phi d\phi=-\int_{0}^{\pi}\sin^{k}\phi d\phi$.
If $k$ is even, $\int_{0}^{2\pi}\sin^{k}\phi d\phi$ is well-known
(by reduction method and integration-by-part). For, let $I_{k}=\int_{0}^{2\pi}\sin^{k}\phi d\phi$,
then for $n\geq1$,
\begin{eqnarray*}
I_{2n} & = & \int_{0}^{2\pi}-\sin^{2n-1}\phi d\cos\phi\\
& = & -\sin^{2n-1}\phi\cos\phi|_{0}^{2\pi}+\int_{0}^{2\pi}(2n-1)\sin^{2n-2}\phi\cos^{2}\phi d\phi\\
& = & (2n-1)\int_{0}^{2\pi}\left(\sin^{2n-2}\phi-\sin^{2n}\phi\right)d\phi\\
& = & (2n-1)(I_{2n-2}-I_{2n}).
\end{eqnarray*}
Hence $I_{2n}=\frac{2n-1}{2n}I_{2n-2}$.
