My question is at the very end of this post, in the final three paragraphs. I want to first lay out the context for the question, so that when it is stated perhaps it is a little clearer why I am asking it.
The substitution formula for integration tells us that if $f'$ and $g'$ are continuous then
$$\int_{g(a)}^{g(b)}f=\int_a^b f(g(x))g'(x)dx\tag{1}$$
The integrand is the derivative of a function $h(x)=F(g(x))$, where $F'=f$. Thus we are integrating $h'(x)=f(g(x))g'(x)$ from $a$ to $b$. This is just $h(b)-h(a)$, which equals $F(g(b))-F(g(a))$, which is $\int_{g(a)}^{g(b)} f$.
It's not a surprising result.
If we think about it in the terms above, then solving an integral such as
$$\int_a^b \sin^5{x}\cos{x}dx\tag{2}$$
is simply a question of identifying the integrand as $f(g(x))g'(x)$.
Thus, $f(x)=x^5$, $g(x)=\sin{x}$, and
$$\sin^5{x}\cos{x}=f(g(x))g'(x)$$
Ie, $\sin^5{x}\cos{x}$ is the derivative of the function $h(x)=F(\sin{x})$, where $F'(x)=f(x)=x^5$.
Therefore, according to the substitution formula, our original integral is the same as
$$\int_{\sin{a}}^{\sin{b}} f$$
$$\int_{\sin{a}}^{\sin{b}} x^5dx$$
$$\left . \frac{x^6}{6} \right |_{\sin{a}}^{\sin{b}}$$
Now let's talk about the shortcuts used to expedite the process of using the substitution formula.
$(1)$ is usually written
$$\int_{g(a)}^{g(b)}f(u)du=\int_a^b f(g(x))g'(x)dx\tag{3}$$
In our calculations above we replaced $(2)$ with
$$\int_{g(a)}^{g(b)} u^5 du$$
Let's forget about the integration limits. They only come up at the end of the computation and the change when we use the substitution rule is the same whether we use shortcuts or not.
So we go from
$$\int \sin^5{x}\cos{x}dx\tag{4}$$
to
$$\int u^5 du$$
Once we identify $g(x)$ we call it $u$. Since $g(x)$ occurs in $f(g(x))$, by replacing the occurrences of $g(x)$ with $u$, we effectively obtain $f(u)$, which is what we want to integrate according to the substitution formula.
The rest of $(4)$, ie the $g'(x)dx$ gets replaced with $du$.
So, my question is this: the shortcuts are simply what I explained above, correct? Just a way to get to $\int f(u)du$ in a way that is easy to remember. There is no actual relationship between $dx$ and $du$?
Okay, I think some people might say "yes, there is a relationship", but if you answer with a yes like this, isn't that because you are leaving the context of integration and using some other part of calculus that is not contained within this question?
At this point in the book (Chapter 19) there has been as of yet no mention of differentials. Hence, I just want to make sure that if you consider this fact (that differentials haven't even been mentioned), then there is no relationship between $dx$ and $du$ other than in a shortcut/trick to obtain the lefthand side of $(3)$ given the righthand side.
