[This integral arises from some script about Jakob Bernoullis solution for the brachistochrone problem]
Let $q: [0,b] \rightarrow \mathbb{R}$ continuous, and the corresponding curve (the slide where the moving object is placed) is $(t,q(t))$ The arc length of this curve is $s(t) = \int_{0}^{b} \sqrt{1+(\dot{q}(\tau))^2} d\tau$.
The "time of fall" is given by
$\int_{0}^{s(b)} \frac{ds}{\sqrt{2gq}} = \int_{0}^{b} \frac{\sqrt{1+\dot{q}^2}}{\sqrt{2gq}} dt$,
where $\sqrt{2gq}$, with $g=$ gravitational constant, is the speed of the object by newtons law of conservation of energy.
But how does this substition work mathematically ? I know that I can put $\frac{ds}{dt} = \sqrt{1+\dot{q}}$ but I would like to do the integration using some explicit function composition $\int f(g(x))g'(x) dx = \int f(u) du$.
$g'(x)$ corresponds to $\sqrt{1+\dot{q}}$, but where is $f(g(x))$ ?
My Idea: I figured to see the first integral as $\int_{0}^{s(b)} \frac{ds}{\sqrt{2gq}} = \int_{0}^{s(b)} \frac{du}{\sqrt{2gq(s^{-1}(u))}}$ which would really yield $\int_{0}^{b} \frac{\sqrt{1+\dot{q(t)}^2}}{\sqrt{2gq(t)}} dt$ on RHS. $s^{-1}$ exists because s is strictly increasing. Furthermore $\sqrt{2gq(s^{-1}(u))}$ gives us the speed at point on the curve with arclength s(u) and this we want to use for calculation, don't we ?
I know that this question is a bit petty, but as a newbie to maths it just bothers me that everyone uses those Leibnitz symbols in fractions and so on but I am not as far in maths to really understand what those mean. I often get the answer "this is not 100% correct but you will see in a few semesters why it works." Still, there has to be a mathematically correct answer to the question above.