Indefinite integral $\int \frac{1}{2-\cos(x)}\,dx$ has discontinuities. How to fix? Using the standard tangent half-angle substitution, we get $$\int \frac{1}{2-\cos(x)}\,dx = \frac{2}{\sqrt{3}}\tan^{-1}(\sqrt{3}\tan\frac{x}{2}) + C$$
The resulting antiderivatives are piecewise continuous functions with discontinuities at $x = (2k+1)\pi$. However, the integrand is continuous everywhere, so any antiderivative must be continuous everywhere (??) according to FTC. So, how do we deal with this problem? What is the correct form of the antiderivatives? Is there a form that avoids the discontinuities?

Your problem is the $+C$ term. Take for example $\frac{1}{x}$. Its most general antiderivative is usually given as
$$\int \frac{1}{x}dx = \ln|x| + C$$
But this is not correct. Notice that the piecewise function 
$$f(x)=\begin{cases} \ln(-x) +2 & x < 0\\ \ln(x) -1 & x> 0\\ \end{cases}$$
is also an antiderivative of $\frac{1}{x}$, even thought the $+C$'s on both sides were different. So any time your integrand has a singularity, the integration constant is allowed to change when you cross it.
On the domain when you integrated, the $+C$ are only uniform between the discontinuities. Otherwise, each discontinuous piece must have its own $+C$. And one can choose a series of $+C$'s such that the resultant antiderivative is continuous.

$\mathbf{EDIT}$: In this case, the continuous antiderivative should be of the form:
$$\frac{2}{\sqrt{3}}\tan^{-1}\left(\sqrt{3}\tan\left(\frac{x}{2}\right)\right) + \frac{2\pi}{\sqrt{3}}\Bigr\lfloor \frac{1}{2\pi}x+\frac{1}{2}\Bigr\rfloor + C$$
This function is both continuous and differentiable at its former discontinuities, even though it may not look like it at first glance.
And a graph of the fixed function, courtesy of Desmos:

