# solving exponential of trigonometric function inside an integral.

I am stuck in the middle while trying to solve this integral.

$\int_{0}^{2\pi} exp({{\sqrt{r}}. (k1 \cos \theta+k2 \sin \theta)}) d \theta$ , where k1 and k2 are known constants, not equal to zero. I tried $\tan \theta =t$ substitution, but the limits became zero to zero. Any help is appreciated. I want a closed form expression in terms of 'r'.

• In the future, when you encounter an integral where the tangent or tangent half angle substitution causes the limits of integration to be equal, you can break the integral up like so $$\int_0^{2\pi} = \int_0^\pi + \int_\pi^{2\pi}$$. Now it's possible to use the tangent half-angle substitution without limits of integration problems. – CodeLabMaster Jan 23 '17 at 11:23

Write $k_1 \cos(\theta) + k_2 \sin(\theta) = \sqrt{k_1^2 + k_2^2} \; \cos(\theta + \alpha)$ for suitable constant $\alpha$. Then $u = \theta + \alpha$ and periodicity makes the integral
$$\int_0^{2\pi} \exp(c \cos(u))\; du$$
where $c = \sqrt{r(k_1^2 + k_2^2)}$. And this is "well-known" to be $2 \pi I_0(c)$ where $I_0$ is a modified Bessel function of the first kind.