Why does Desmos plot $\frac1{50}(-\ln(\cot(\frac{x}{2}))(\cos x-1)+\frac12x^2-2\ln(\cos(\frac{x}{2})))$ as a dotted parabola? $$\frac{-\ln\left(\cot\left(\dfrac{x}{2}\right)\right)\left(\cos\left(x\right)-1\right)+\dfrac{x^2}{2}-2\ln\left(\cos\left(\dfrac{x}{2}\right)\right)}{50}$$
https://www.desmos.com/calculator/u2kmp0y6ju
If you see the graph in the above link, it will appear as a dotted parabola.
Could someone explain what part of the function makes it appear like this.
Also are there other functions like this that create dotted parabolas? 
Edit: Turns out this function has some pretty interesting graphs. If you are interested, here are some interesting plots that I got when I played around with the function:
https://www.desmos.com/calculator/bio8nxkrry
 A: The natural log of a negative number does not exist (as a real number), so your function is undefined for all values of $x$ that would yield the expression $\ln(a)$ for some $a \leq 0$. 
Since your function includes $\ln(\cos(x/2))$, we need to find where $\cos(x/2)\leq 0$, and your function will be undefined there. This happens for all $x \in [(4k+1)\pi, (4k+3)\pi]$ for some integer $k$.
Your function also includes $\ln(\cot(x/2))$, so we need to find where $\cot(x/2) \leq 0$. This happens for all $x \in [(4k+1)\pi, (4k+2)\pi] \cup [(4k+3)\pi, (4k+4)\pi]$ for some integer $k$.
So your function is only defined for $x \in (4k\pi,(4k+1)\pi)$ for some integer $k$. This is why we see a dashed line.
As for why it is a parabola, we can re-write the equation as
$$\frac{x^2}{100} + \frac{1}{50}\left(-\ln\left(\cot\left(\frac{x}{2}\right)\right)\left(\cos\left(x\right)-1\right)-2\ln\left(\cos\left(\frac{x}{2}\right)\right)\right).$$
It turns out that everything after the $\frac{x^2}{100}$ is close to zero (but not exactly zero) when it exists (try graphing it by itself: https://www.desmos.com/calculator/izmnkhpjae). So you could actually add this function to any function and see a dotted line close to the original function. For instance: https://www.desmos.com/calculator/9gvrd5dkzl.
A: The $\log$ function introduces a domain restriction. Recall only strictly positive numbers can go in that function.
