Evaluating definite integrals using Fundamental Theorem of Calculus

A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.

@Martin Argerami posted a succinct and correct answer:

A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.

To elaborate, $F_2$ is not continuous due to the multi-valued nature of $\arctan$. Since $\tan$ is a periodic function (with a period of $\pi$), this means that $\arctan$ has an infinite number of values. Its plot looks like this:

To make it a proper function (one unique output value), we usually pick the primary branch (the one through the origin). However, notice that the primary branch is discontinuous at infinity - if we pass through positive infinity and come back out at negative infinity, we've jumped from $\pi/2$ to $-\pi/2$. What we really wanted was to move to the next branch up to preserve continuity, but to do that we'd no longer have a single valued function. We wouldn't have a function at all, really, since the common definition of function requires it to be single valued.

The discontinuities of $F_2$ are precisely where the argument of $\arctan$ becomes infinite - where $2+\cot\frac{x}{2}$ and $2+\tan\frac{x}{2}$ become zero - approximately 4.07 and 5.36.