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?

$$\frac{1}{\sqrt{R^2+x^2-2 R x \cos (\theta )}}=\frac{1}{\sqrt{R^2+x^2}}\frac{1}{\sqrt{1-k\cos (\theta )}}$$ 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
$$\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)=2\pi\frac{1-\frac{497 }{576}k^2+\frac{3835}{36864}k^4 } {1-\frac{605 }{576}k^2+\frac{7315 }{36864}k^4 }$$
