I am trying to prove substitution of integrals by going from the chain rule. Here is my attempt:
$$\frac{d}{dx}f(u)=\frac{d}{du}f\frac{d}{dx}u$$
$$\int \frac{d}{dx}f(u)dx=\int \frac{d}{du}f \frac{d}{dx}u dx=\int \frac{df}{du} du $$
In other words what we need is the core $\frac{du}{dx}dx=du$
So by taking $\frac{du}{dx}dx$ out and replacing it with du we have arrived at a new integral:
$$\int \frac{df}{du} du $$
Is this correct? It seems that we are using the chain rule to arrive at this result?
It all depends on the regularity of your functions. Lets say that $f$ is continuous on some interval $I=]a,b[$ and that $g$ is continuously differentiable on $J=]c,d[$ with values in $I$.
Now since $f$ is continuous it has an antiderivative $F$ such that $F'(u)=f(u)$ for all $u \in I$. Now lets define a function $h\colon J \to \mathbb R$ by $h(x) = F(g(x))$. Using the chain rule we get that: \begin{equation} h'(x) = F'(g(x))g'(x) = f(g(x))g'(x). \end{equation} Thus $h$ is an antiderivative for $f(g(x))g'(x)$. Using the notation of indefinite integrals we have that: \begin{equation} \int f(g(x))g'(x)dx = F(g(x)) = \int f(u) du_{|u=g(x)}. \end{equation}
So in essence your argument is on the right track.