Integration $\int\frac{\mathrm{d}v}{\mathrm{d}t}\,\mathrm{d}x=\int\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}v$ I saw this in a proof of the formula for KE. $$\int\frac{\mathrm{d}v}{\mathrm{d}t}\,\mathrm{d}x=\int\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}v$$
Why can you just swap the two differentials? I can't find a proof anywhere.
 A: As Cameron alluded to, implicit in the question statement are a number of smoothness and invertibility  assumptions.
The underlying tool is the change of variables theorem which states that
$$ \int_{\phi(x_0)}^{\phi(x_1 )} f(t) dt = \int_{x_0}^{x_1} f(\phi(x)) \phi'(x) dx $$
(again with appropriate smoothness requirements and $\phi$ being suitable monotonic).
Since $x,v$ are functions of $t$, we start with
$I=\int_{t_0}^{t_1} v'(t)x'(t) dt$.
If we assume that we have some $\phi$ such that $x(\phi(\xi)) = \xi$, then we see that $x'(\phi(\xi)) \phi'(\xi) = 1$. Letting $t_0 = \phi(x_0)$, $t_1 = \phi(x_1)$, the change of variables theorem gives
$I = \int_{\phi(x_0)}^{\phi(x_1 )} v'(t)x'(t) dt = \int_{x_0}^{x_1} v'( \phi(x) )x'( \phi(x) )  \phi'(x) dx  $, which gives
$I = \int_{x_0}^{x_1} v'( \phi(x) ) dx  $.
If it is also true that for some suitable monotonic $\eta$ that $v(\eta(\zeta)) = \zeta$, then a similar analysis yields $I = \int_{v_0}^{v_1} x'( \eta(\nu) ) d\nu  $, with $t_0 = \eta(v_0)$, $t_1 = \eta(v_1)$.
Equating the two gives the formula in the question, albeit somewhat more long winded.
A: By the chain rule, we have
$$\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt} \\\\\Rightarrow \frac{dv}{dt}dx =  \frac{dx}{dt} dv$$
Your result is obtained by integrating both sides of this expression. As the above answer said, this is merely a heuristic process.
A: If you view $x$ as a function of $t$, you have that $dx = \frac{dx}{dt} dt$ and so
$$\int \frac{dv}{dt}dx = \int \frac{dv}{dt}\frac{dx}{dt}dt.$$
Then by similar reasoning, if you view $t$ as a function of $v$, you have that $dt = \frac{dt}{dv} dv = \left(\frac{dv}{dt}\right)^{-1}dv$ by the inverse function theorem and so
$$\int \frac{dv}{dt}dx = \int\frac{dx}{dt}dv.$$
Physicists do this a lot. They treat differentials as fractions and cancel them willy-nilly. While the answer is generally correct, what is really going on is chain rule married with the inverse function theorem oft times.
