Evaluate $\int_0^{2\pi}{\exp(A \cos(x))dx}, \int_0^{2\pi}{\exp(A \cos(x) + B \sin(x))dx}$

Question

I saw that $$\int_0^\pi{\exp(A \cos(x))dx} = \pi I_0(A) $$

I am trying to understand what would be the solution for the following two integrals :

$$ \begin{split} &\int_0^{2\pi} \exp\left(A \cos(x)\right)dx \\ &\int_0^{2\pi} \exp\left(A \cos(x) + B \sin(x)\right)dx \end{split} $$

I anticipate that this might come interms of Bessels functions, I am clueless on how to proceed with the known solution. Any help is appreciated.

Answer 1

First, remember SolubleFish's comment: $$\int_0^{2\pi}\exp(s\cos t)\mathrm dt=2\pi I_0(s)$$ Now, note that $$x\cos t+y\sin t=\sqrt{x^2+y^2}~\cos\left(t-\arctan(y/x)\right)$$ (proof) therefore $$\int_0^{2\pi}\exp(x\cos t +y\sin t)\mathrm dt \\ =\int_0^{2\pi}\exp\left(\sqrt{x^2+y^2}~\cos(t-\arctan(y/x))\right)\mathrm dt$$ But due to the $2\pi$ periodicity of the integrand, we can replace $t-...$ by just $t$: $$\int_0^{2\pi}\exp(x\cos t+y\sin t)\mathrm dt=\int_0^{2\pi} \exp\left(\sqrt{x^2+y^2}~\cos t\right)\mathrm dt=2\pi I_0\left(\sqrt{x^2+y^2}\right)$$ Done.

Answer 2

$Acos(x)+Bsin(x)=Ccos(x-a)$. Get $C$ from $Ccos(a)=A$ and $Csin(a)=B$. Second integral=$2\pi(I_0(C))$.