Problem.
A particle moves in a deaccelerated manner, describing a circular trajectory of radius $r$, having an initial speed $v_0$. Suppose $a_n=-a_t$ (normal acceleration and tangential acceleration).
Calculate the speed of the particle when it's travelled an arc length of $3r$.
Solution.
Remember $a_n = \frac {v^2}r$ and $a_t= v'$ (where $v=|\vec v|$).
Then $\frac {dv}{dt}=-\frac {v^2}{r}$, but $\frac {dv}{dt}=\frac {dv}{ds}\frac {ds}{dt}$. Recall that $s(t)=\int_0^t v(t)dt$, so $\frac {ds}{dt}=v(t)$.
Then, $\frac {dv}{ds}=-\frac {v}{r}$, and $\frac {dv}{v}=-\frac {ds}{r}$.
We now integrate: $$ \int_{v_0}^v \frac {dv}{v}=-\frac 1 r\int_0^{3r}ds $$
To reach $v=v_0e^{-3}$.
I understand the separation of variables, as I'm used to that from DEs, what I don't understand is why he can integrate on one side with the limits $v_0,v$ ($v$ is also an integration variable!!!) and on the other side using $0,3r$.
Why is this possible?
Is there a more mathematically correct way to solve this? What about the part saying $s'(t)=v(t)$ (they've justified that using the FTC, but here we have $t$ as both a limit of integration and dummy variable...)?
Here's what's really happening. We know $$\frac{1}{v}\frac{dv}{ds}=-\frac{1}{r}.$$ Now we integrate both sides with respect to $s$. $$\int_0^{3r}\frac{1}{v}\frac{dv}{ds}ds=-\frac{1}{r}\int_0^{3r}\,ds.$$ We solve the left hand side by realizing that $v$ is really a function of $s$, $v=v(s)$. Then $dv=\frac{dv}{ds}ds$. So we can make the substitution in the integral. Of course, this means that we also have to substitute the endpoints. When $s=0$, $v(s)=v_0$. When $s=3r$, $v(s)=v$ (again, as you note, we are being a little sloppy with the notation here.) The integral becomes $$\int_{v_0}^v\frac{dv}{v}=-\frac{1}{r}\int_0^{3r}\,ds.$$
This method of solving problems is so commonplace in physics that physicists often abuse notation and treat $\frac{dy}{dx}$ as a fraction. While not a proper treatment of the derivative, it gets the job done.