The indefinite integral has no known closed form, but it is known that:
$$\int_{-\pi}^{\pi} e^{i \xi \cos(x)} dx = 2 \pi J_0(\xi)$$
Where $J_0$ is a Bessel function of the first kind. Is there a way to obtain a solution if you add a (real) offset, $a$, to the integration boundaries?
$$\int_{-\pi + a}^{\pi + a} e^{i \xi \cos(x)} dx$$
Your integral also has the value $2\pi J_0(x)$. To see this, let us look at the more general case $$I(\theta_0)=\int_{-\pi+\theta_0}^{\pi+\theta_0}\mathrm{d}\theta\,f(\cos(\theta))$$ for any "nice" function $f(x)$. Rearranging the integral gets us $$I(\theta_0)=\int_{-\pi}^{\pi}\mathrm{d}\theta\,f(\cos(\theta-\theta_0)),$$ which after taking the derivative w.r.t. $\theta_0$ gives us $$\frac{\mathrm{d}}{\mathrm{d}\theta_0}I(\theta_0)=\int_{-\pi}^{\pi}\mathrm{d}\theta\,\frac{\partial}{\partial\theta_0}f(\cos(\theta-\theta_0))=\int_{-\pi}^{\pi}\mathrm{d}\theta\,f'(\cos(\theta-\theta_0))\sin(\theta-\theta_0).$$ Looking at the integrand, it's the same as $$f'(\cos(\theta-\theta_0))\sin(\theta-\theta_0)=-\frac{\mathrm{d}}{\mathrm{d}\theta}f(\cos(\theta-\theta_0)),$$ therefore we get $$\frac{\mathrm{d}}{\mathrm{d}\theta_0}I(\theta_0)=-f(\cos(\theta-\theta_0))\Big|_{-\pi}^\pi=0,$$ because $\cos(\pi-\theta_0)=\cos(-\pi-\theta_0)\implies f(\cos(-\pi-\theta_0))=f(\cos(\pi-\theta_0))$. From this follows then, that $I(\theta_0)$ has to be a constant and thus $$\boxed{I(\theta_0)=\int_{-\pi+\theta_0}^{\pi+\theta_0}\mathrm{d}\theta\,f(\cos(\theta))=C}$$ So figuring out the value for $I(0)$ gives you the value for all $I(\theta_0)$.