I am trying to derive the Fourier series of the function
$$f(\phi)=\sqrt{1+ a \cos(\phi)}=\sum_{m=0}^{\infty}b_{m}(a)\cos(m\phi)$$
The most obvious approach is to use the standard Fourier coefficient formula:
$$b_{m}(a)=\frac{1}{\pi}\int_{0}^{2\pi}\cos(m\phi)\sqrt{1+ a \cos(\phi)}\,dx$$
However I have no idea how to approach this integral. Instead, I have tried using power series methods. Applying the binomial formula for $\sqrt{1+x}$,
$$\sqrt{1+ a \cos(\phi)}=\sum_{k=0}^{k=\infty}{{1/2}\choose{k}}a^{k}\cos(\phi)^{k}$$
We may derive the following standard formula for $\cos(\phi)^{k}$ using $\cos(\phi)=(e^{i\phi}+e^{-i\phi})/2$ and the binomial expansion for $(x+y)^{k}$,
$$\cos(\phi)^{k}=\frac{1}{2^{k}}\sum_{n=0}^{k}{{k}\choose{n}}e^{i(2n-k)\phi} \\ \begin{aligned}\Rightarrow \sqrt{1+ a \cos(\phi)}&=\sum_{k=0}^{\infty}{{1/2}\choose{k}}\left(\frac{a}{2}\right)^{k}\sum_{n=0}^{k}{{k}\choose{n}}e^{i(2n-k)\phi} \\ &=\sum_{m=-\infty}^{\infty}c_{m}(a)e^{im\phi} \end{aligned}$$
We will consider separately the cases of $m$ even and $m$ odd to compute $c_{m}(a)$.
m even:
$m=2m'$ and $2n-k=2m'$ implies $k=2k'$. Therefore $n=k'+m'$ and
$$ c_{2m'}(a)=\sum_{k'=0}^{\infty}\left(\frac{a}{2}\right)^{2k'}{{1/2}\choose{2k'}}{{2k'}\choose{k'+m'}}$$
m odd:
$m=2m'+1$ and $2n-k=2m'+1$ implies $k=2k'+1$. Therefore $n=k'+m'+1$ and
$$c_{2m'+1}(a)=\sum_{k'=0}^{\infty}\left(\frac{a}{2}\right)^{2k'+1}{{1/2}\choose{2k'+1}}{{2k'+1}\choose{k'+m'+1}}$$
Again using the complex exponential formula for $\cos(\phi)$, we may relate $b_{m}(a)$ to $c_{m}(a)$. In particular, $b_{0}(a)=c_{0}(a)$ and $b_{m}(a)=2c_{m}(a)$ for $m\neq 0$.
Thus my question reduces to the following; is there any closed form expression for $c_{m}(a)$, using power series methods or otherwise? I realise I might have been overly hasty in dismissing the integration route.