Why is $ \ \int^x_{x_0} v\frac{dv}{dx} \, dx =\int^v_{v_0} v \, dv$? Why is $$\int^x_{x_0} v\frac{dv}{dx} \, dx =\int^v_{v_0} v \, dv$$
where $v$ is a function of $x$, and $x$ is a function of time $t$?
I know it involves some substitution, but which one?   The only thing I see is letting $\ dx=dv/\frac{dv}{dx} \ $ but why are we allowed to "just define" $dx$ to equal that? 

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 A: This is a substitution problem. You have gone from an $x$ variable to $ v$ variable. Limits of integration change as a result. As $ x $ goes from $ x_0 $  to $x_1$ , $v$ goes from $v_0$ to $v_1$.  
A: First let us be clear on the meanings of "free variable" and "bound variable".
$$
\sum_{j=1}^4 j^2 = 1^2 + 2^2 + 3^2 + 4^2 = \sum_{k=1}^4 k^2.
$$
The variables $j$ and $k$ above are bound variables. One can freely change the name of a bound variable from $j$ to $k.$ The value of the sum $1^2 + 2^2 + 3^2 + 4^2$ does not depend on the value of anything called $j$ or $k.$ On the other hand the value of
$$
\sum_{j=1}^4 \cos(j\ell)
$$
depends on what number $\ell$ is. So $\ell$ is a free variable in this expression.
The chain rule says that if $h(x) = g(v(x))$ then $h'(x) = f'(v(x))v'(x).$ Therefore
$$
\int_a^b f'(v(x))v'(x)\, dx =\int_a^b h'(x) \, dx = h(b) - h(a) = \underbrace{f(v(b)) - f(v(a)) = \int_{v(a)}^{v(b)} f'(v) \, dv}.
$$
In the integral over the $\underbrace{\text{underbrace}},$ $v$ is a bound variable and we can rename it freely. We could just as correctly have written
$$
f(v(b)) - f(v(a)) = \int_{v(b)}^{v(a)} f'(w)\, dw.
$$
We are not "just defining" $dx$ that way; we are "just defining" $dv$ that way, and we could have equally correctly called in $dw.$
What justifies integration by substitution is the chain rule. The chain rule is differentiation by substitution. For example in
$$
y = (x^3 - 5x+ 12)^{40},
$$
one can use the substitution $u = x^3 - 5x+12$ and then write
$$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 40u^{39} \cdot (3x^2 - 5) = 40(x^3 - 5x + 9)^{39}\cdot (3x^2 - 5).
$$
A: The notation here is atrocious; presumably what is meant on the left is
$$\int_{x_0}^{x_1} v(x) \frac{dv}{dx}(x)\,dx,$$
with $v$ a diffeomorphism on $(x_0,x_1)$. We can introduce the new variable $y = v(x);\ dy = \frac{dv}{dx}dx$ and make the substitution
$$\int_{x_0}^{x_1} v(x) \frac{dv}{dx}(x)\,dx = \int_{v(x_0)}^{v(x_1)} y \,dy.$$
