Anti-derivative of continuous function $\frac{1}{2+\sin x}$ I use tangent half-angle substitution to calculate this indefinite integral:
$$
\int \frac{1}{2+\sin x}\,dx = \frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}+\text{constant}.
$$
Wolfram Alpha also give the same answer. However, $\frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}$ is discontinuous on $(n+1)\pi$ where $n$ is any integer. Why is an anti-derivative of a continuous function discontinuous?
 A: Let's examine the first troublesome positive point, that is, $\pi$. We know that an antiderivative in the interval $(-\pi,\pi)$ is
$$
f_0(x)=\frac{2}{\sqrt{3}}\arctan\frac{2\tan(x/2)+1}{\sqrt{3}}+c_0
$$
We also know that an antiderivative in the interval $(\pi,3\pi)$ is of the form
$$
f_1(x)=\frac{2}{\sqrt{3}}\arctan\frac{2\tan(x/2)+1}{\sqrt{3}}+c_1
$$
Note that
$$
\lim_{x\to\pi^{-}}f_0(x)=\frac{\pi}{\sqrt{3}}+c_0
$$
and
$$
\lim_{x\to\pi^{+}}f_1(x)=-\frac{\pi}{\sqrt{3}}+c_1
$$
so in order to get continuity at $\pi$ we have
$$
c_1=\frac{2\pi}{\sqrt{3}}+c_0
$$
Do the same for the other intervals.
A: One thing not emphasized much in the conventional calculus curriculum is that things like
$$
\int \frac{dx} x = \log|x| + \text{“constant”}
$$
are not true unless one takes “constant” to mean piecewise constant:
$$
\int \frac{dx} x = \log|x| + \begin{cases} \text{one constant} & \text{if }x>0, \\ \text{another constant} & \text{if }x<0. \end{cases}
$$
and:
\begin{align}
& \int \sec x\,dx \\[4pt]
= {} & \log|\sec x+\tan x| + \cdots \text{what?} \\
& \cdots + \text{a different constant on each interval between vertical asymptotes.}
\end{align}
The comments under the question itself are pretty good so far:


*

*“egreg” points out that the technique involving the tangent half-angle substitution is valid only on intervals between vertical asymptotes of the function $x\mapsto\tan\frac x 2$. That means it doesn't rule out anything that happens at those points: it doesn't say that there's an answer there or that there's not.

*Jeppe Stig Nielsen points out that the antiderivative must be everywhere increasing since the function being integrated is everywhere positive.  That means the answer cannot be a periodic function.

*“runaround” and “KCd” remind us that there is such a thing as removable discontinuities.

*You yourself point out that the antiderivative of a continuous function should be continuous.


Now just put all four of these points together and figure out which “piecewise constant” will give you a continuous function.  That function will be everywhere increasing.
A: We have the following definite integral:
$$f(x)=\color{#dd1111}{\int_0^x\frac{dt}{2+\sin(t)}}\stackrel?=\color{#4488dd}{\frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}}-\frac\pi{3\sqrt3}$$
At first, I wouldn't doubt this solution:

But then I look at the big picture:

It's not that the integral in question is discontinuous, it's that
a) the proposed solution is discontinuous for fixed constant.
b) a u-substitution was made that was not valid for $|x|\ge\pi$.  To remove this problem, one could take $n=x\text{ mod }2\pi$ so that we make the argument between $(-\pi,\pi)$ and add in a linear piece of $n\int_{-\pi}^{\pi}\frac{dt}{2+\sin(t)}$.
