I am trying to learn about differential forms. I think I understand what they are and the basic idea of integrals. But I wanted to make sure that I understand the process of integration by substitution using differential forms.
My basic question is how to do integration by substitution using differential forms.
I have tried the following example:
Say I want to calculate $$ \int_1^2 x\sqrt{x^2 +1}\; dx. $$ If I can recognize $f(x) = 2x\sqrt{x^2 + 1}\; dx$ as the differential of a $0$-form, then I can use Stoke's theorem. So I define the $0$-form $u = x^2 + 1$ and find the differential $du = 2x\;dx$. That is $\frac{1}{2}u = x\; dx$. So these two $0$-forms are the same as $1$-forms. Then as $1$-forms
$$\frac{1}{2}\sqrt{u}\;du = x\sqrt{x^2+ 1}\;dx$$
Then I am integrating using Stoke's Theorem $$\begin{align} \int_1^2 x\sqrt{x^2 +1}\; dx &= \int_1^2 \frac{1}{2}\sqrt{u}\;du = ... \end{align} $$ Is that saying it the right way? Do I have to change the limits on the integral? How does that work?
You have the integral $$\int_1^2 x\sqrt{x^2 +1}\; dx$$
We can consider this to be the integral of the one-form $f(x)\ dx = x\sqrt{x^2 +1}\; dx$ on the "$1$-dimensional manifold with boundary" $M=[1,2]$.
The way that "pulling back" and "pushing forward" works in differential forms is summed up in this formula: $$\int_{N} \omega = \int_{M} \phi^{*}\omega$$
where we define the pullback of $\omega$ by $\phi$ as $\phi^* \omega := \omega \circ \phi$ and $M$ and $N:=\phi(M)$ are the regions of integration.
Sometimes the way to simplify our integrand will be to recognize it as the composition of a differential form with another function (like in this example). That's called a "pushfoward" and it's described by starting from the RHS of the above formula and finding the simpler LHS. Other times we'll want to compose our differential form with another function to simplify it. That's called a "pullback" and it's described by starting from the LHS of the above formula and finding the simpler RHS.
Here we'd like to use a pushforward. So essentially we just need to find a bijective, differentiable function $\phi: M \to N$ that we can use to simplify the integrand. You make a good suggestion in your post with $u=x^2+1$ (notice that this is bijective on $[1,2]$). Let's let $\phi(x) = x^2+1$. Then we can see that if $\omega$ is our simplified differential form, then $$f(x)\ dx = \phi^* \omega = \omega\circ \phi = g(\phi(x))\phi'(x)\ dx = x\sqrt{x^2+1}\ dx$$
Then, knowing that $d\phi = \phi'(x)\ dx = 2x\ dx$, we see that clearly $g(t)\ dt = \frac 12\sqrt{t}\ dt = \omega$ (do you see where the $\frac 12$ comes from?).
So then $$\int_1^2 x\sqrt{x^2+1}\ dx = \int_{M} \phi^*\omega = \int_{\phi(M)} \omega = \int_{\phi(1)}^{\phi(2)} \frac 12\sqrt{t}\ dt = \int_1^5 \frac 12\sqrt{t}\ dt$$
This last integral is much easier to solve (using the fundamental theorem of calculus; AKA Stokes' theorem).
Notice that this is exactly the same procedure as you would use with regular scalar calculus (only you call it $u$ instead of $\phi$). In a sense, scalar calculus is already "complete". By that I mean that all of those cool generalizations we've got for it, like integration of differential forms and tensors, don't really add anything to the $1$-dimensional case. To really see where differential forms are useful you'll have to wait until you start studying more interesting manifolds in differential geometry.
For an example of a pullback, consider the integral $$\int_0^1 \frac{1}{\sqrt{1-x^2}}\ dx$$
I claim that composing the differential form $\frac{1}{\sqrt{1-x^2}}\ dx$ with the function $\phi(\theta) = \sin(\theta)$ will make this integral easier to evaluate. Give it a try.
Hint: when starting from the LHS of the formula I give above you'll find that $M=\phi^{-1}(N)$ -- this is why $\phi$ has to be a bijective function. Is $\sin(\theta)$ a bijective function on $N=[0,1]$?