I'm solving an integral in three coordinates. One of the coordinates is the integral: $\int_{0}^{2\pi} e^{\sin(x)+\cos(x)} dx$.
Is possible to get a closed form for that?
My efforts.
Let $u=-x$.
$\int_{0}^{2\pi} e^{\sin(x)+\cos(x)} dx =-\int_{0}^{2\pi} e^{\sqrt{2}\sin(\pi/4-u)} du $
Let $$ I=\int_0^{2\pi}e^{\sin(x)+\cos(x)}\,dx. $$ From the trigonometric addition formula $\cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$, with $\alpha=\pi/4, \beta=x$, using the special value $\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt2$, we have $$ \sin(x) + \cos(x) = \sqrt2\cos\left(\pi/4-x\right). $$ From here we have \begin{align} I &= \int_0^{2\pi}e^{\sqrt2\cos\left(\pi/4-x\right)}\,dx\\ &= \int_0^{2\pi}e^{\sqrt2\cos\left(x\right)}\,dx \tag{1}\\ &= \int_0^{\pi}e^{\sqrt2\cos\left(x\right)}\,dx + \int_\pi^{2\pi}e^{\sqrt2\cos\left(x\right)}\,dx \tag{2}\\ &= \int_0^{\pi}e^{\sqrt2\cos\left(x\right)}\,dx + \int_0^{\pi}e^{-\sqrt2\cos\left(x\right)}\,dx. \tag{3}\\ \end{align} In $(1)$ we used the substitution $x\mapsto\pi/4-x$ and the periodicity of the integrand. In $(2)$ we used the additivity of integration on intervals. In $(3)$ for the second integral we used the substitution $x \mapsto x-\pi$, and the symmetry of the cosine function, $\cos(x-\pi) = -\cos(x)$.
Using the definition and the properties of the modified Bessel function of the first kind, we have $$ I_0(z) = \frac{1}{\pi}\int_0^\pi e^{z\cos(t)} \,dt = \frac{1}{\pi}\int_0^\pi e^{-z\cos(t)} \,dt. $$ Putting all this together, we have $$ I = 2\pi I_0\left(\sqrt2\right). $$