Can $\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$ where $R$ and $x$ are positive constants, be solved using substitution? While I was finding the potential at a point in the plane of a uniformly charged ring, I got the following integral as the solution.
$$\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$$
where $R$ and $x$ are positive constants.
Can this integral be solved using substitution?
 A: $$\frac{1}{\sqrt{R^2+x^2-2 R x \cos (\theta )}}=\frac{1}{\sqrt{R^2+x^2}}\frac{1}{\sqrt{1-k\cos (\theta )}}$$ with $k=\frac{2Rx}{R^2+x^2}$.
$$\int \frac{d\theta}{\sqrt{1-k\cos (\theta )}}=\frac 2 {\sqrt{{1-k}}} F\left(\frac{\theta }{2}|\frac{2 k}{k-1}\right)$$
$$\int_0^{2\pi} \frac{d\theta}{\sqrt{1-k\cos (\theta )}}=\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)$$
If $k$ is small, you can use the expansion
$$\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)=2\pi \left(1+\frac{3 k^2}{16}+\frac{105 k^4}{1024}+\frac{1155 k^6}{16384} \right)+O\left(k^8\right)$$
If you want a much better approximation, you could use the Padé approximant
$$\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)=2\pi\frac{1-\frac{497 }{576}k^2+\frac{3835}{36864}k^4 } {1-\frac{605 }{576}k^2+\frac{7315 }{36864}k^4 }$$
A: Once even I tried to derive this result and got stuck on the same step. After searching many articles and abstracts on internet, I concluded that there is no simple analytical solution of this integral rather a lot of heavy computation is required.
The potential equation was like $$V=\int_0^{2\pi}k\cdot\frac{Q\ d\theta}{2\pi}\cdot\frac{1}{\sqrt{R^2+x^2-2Rx\cos\theta}}=\frac{k\ Q}{\pi\ R}\int_0^{\pi}\frac{d\theta}{\sqrt{a-b\cos\theta}}$$
where  $\displaystyle a=1+\frac{x^2}{R^2}$  and  $\displaystyle b=\frac{2x}{R}.$
1. One possibility is to express the integrand as a power series in  $\cosθ$, and then integrate term by term.
Integrate this series, using  $$\int_0^{\pi} \cos^nθ\ dθ=\begin{cases}\frac{(n−1)!!π}{n!!} &, \text{if $n$ is even}\\ 0 &, \text{if $n$ is odd}\end{cases}$$
$$V=\frac{k\ Q}{R}(1+\frac{3}{16}c^2+\frac{105}{1024}c^4+\frac{1155}{16384}c^6+\frac{25025}{4194304}c^8+...)$$
where $\displaystyle c=\frac{b}{a}=\frac{2(x/R)}{(x/R)^2+1}.$
2.  We can obtain a power series in  $x/R$ .
Consider the expression  $\displaystyle \frac{1}{\sqrt{R^2+x^2−2Rx\cosθ}}=\frac{1}{R\sqrt{1+(x/R)^2−2(x/R)\cosθ}}$
Expanding this trinomial, we get a Legendre Polynomial whose co-efficients can be calculated. After integrating, we get,
$$V=\frac{k\ Q}{R}\left(1+\frac{1}{4}(\frac{x}{R})^2+\frac{9}{64}(\frac{x}{R})^4+\frac{25}{256}(\frac{x}{R})^6+\frac{1225}{16384}(\frac{x}{R})^8...)\right)$$
