How to justify the implication $\frac{du}{dx}=\cos x\implies du=\cos x\,dx$ using differential forms?

Question

Consider integral $\int_{0}^{\frac{\pi}{2}}\sin x\cos xdx$, I substitute $u=\sin x$, $\frac{du}{dx}=\cos x$, $du=\cos xdx$.

How can we prove last step, multiplying by $dx$. Everywhere I see everyone says it can be justified by differential forms. But how? Where can I read proof and not just sentence that it can be justified by differential forms?

Answer 1

That operation is the exterior derivative.

Essentially, a smooth function can be understood as a zero form. Then you have the exterior derivative which is an operator that takes $k-$forms into $(k+1)-forms$. In local coordinates (to avoid the generality) if $\omega = \sum_I a_Idx_I$ is a $k-$form where $a_I$ are smooth functions on a manifold, then its exterior derivative is $$d\omega = \sum_{j}\sum_I\frac{\partial a_I}{x_j}dx_j \wedge dx_I.$$

In your case, $f$ is a smooth function, hence a $0-$form on $\mathbb{R}$. So, $df = \frac{\partial f}{\partial x}dx$

Answer 2

Yes, it is a fiction in the notation that $\frac{du}{dx}$ looks like a fraction of numbers, when $du$ and $dx$ are not numbers.

It is a useful fiction - for example, the chain rule can be written as $$\frac{du}{dt}=\frac{du}{dx}\frac{dx}{dt}.$$ There are reasons this fiction works, but it can be tough to explain in an intro class. The question might even be, "What does '$\cos x\,dx$' mean in isolation? What does it mean for it to be equal to $du?$"

And yes, the ultimate answer to that question is in differential forms. But that is a complex topic, which I'll skip here.

In the end, at an intro level, it is better to phrase this result in terms of functions. Treat this as a substitution rule, and prove separately:

If $$F(u)=\int f(u)du\tag1$$ then $$F(g(x))=\int f(g(x))\,g'(x)\,dx\tag 2$$

But $(1)$ means $F'(u)=f(u)$ and $(2)$ means $(F\circ g)'(x)=f(g(x))g'(x).$ And this is just the chain rule.

Your case is $f(u)=u$ and $g(x)=\sin x.$

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An additional notation wrinkle is to use Riemann-Stieljes integrals. We can show, if $u(x)$ is differentiable, then:

$$\int_{a}^{b}g(x)u'(x)\,dx=\int_{a}^{b} g(x)\,du(x),\tag3$$ where the right side is a Riemann-Stieljes integral.

In this sense, $du(x)=u'(x)\,dx$ as distributions.

A distribution is a linear map which takes a function $g$ to a number. In this case, $du(x)$ takes $g$ to $\int_{a}^{b} g(x)du(x)$ and $u'(x)\,dx$ takes $g$ to $\int_{a}^b g(x)u'(x)\,dx.$ And we see from $(1)$ that both are equal.

When $f$ is a function whose domain includes the range of $u,$ then we can treat these as distributions on such $f$ by applying the distributions to $f\circ u.$ Then we get from the first part of this answer:

$$\int_a^b f(u(x))du(x) =\int_{u(a)}^{u(b)} f(u)\,du$$

Ultimately, you can think of Stieljes integrals as an introduction to the idea of general distributions, which will ultimately lead to differentiable forms.