ostensibly discontinuous antiderivative

The real function

$x\mapsto {\displaystyle \frac{1}{5-3\cos x}}$(1)

is continuous for any $x$ (the denominator is always positive) and therefore it has an antiderivative, defined for all $x$.  Using the universal trigonometric substitution

$$\cos x:=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2},t=\tan \frac{x}{2},$$

we obtain

$$5-3\cos x=\frac{5(1+t^2)-3(1-t^2)}{1+t^2}=\frac{2(1+4t^2)}{1+t^2},$$

whence

$$\int \frac{dx}{5-3\cos x}=\int \frac{dt}{1+4t^2}=\frac{1}{2}\arctan 2t+C=\frac{1}{2}\arctan \left(2\tan \frac{x}{2}\right)+C.$$

This result is not defined in the odd multiples of $\pi$, and it seems that the function (1) does not have a continuous antiderivative.

However, one can check that the function

$x\mapsto {\displaystyle \frac{x}{4}}+{\displaystyle \frac{1}{2}}\arctan {\displaystyle \frac{\sin x}{3-\cos x}}+C$(2)

is everywhere continuous and has as its derivative the function (1); one has