Trigonometric substitution and $dx$

Question

Normally, when computing integrals, I tend to see a $dx$ as "a little bit of x", but really rather use it to denote what is a variable, and what is a constant. Most of the answers on this question support that notion.

Now I started looking into trigonometric substitutions, and in various sources a term $dx$ becomes an important part of the expression after substitution. For example, this wikipedia page starts off with the following integral: $$\int{\frac{dx}{\sqrt{a^2-x^2}}}$$

and then defines: $$ x = a(\sin(\theta)),\,dx =a(\cos(\theta))\,d\theta$$

where clearly, $dx$ becomes the derivative of the substitution for $x$ times $d\theta$.

Is this the same $dx$ as in other integrals? Why does it become such a key-role in these substitutions, when in other integrals it is viewed as just a "notational thing" to denote the variable? I would have expected that a change of variables would have just turned a $dx$ into a $d\theta$. Why does it not work that way?

Answer 1

if you really want to understand this, perhaps the only way is to go under the hood and see how a change of variables really works. A COV is not the fact that: $$\int{U(x) dx } \,=\, \int{U(t) dt } \,=\, \int{U(i) di }.$$

This is merely notation; it is picking a letter.

Let's work with $f(x) \,=\, (4-x^2)^{-1/2}$ defined on $(-2,2)$. Because this function is continuous, it is the derivative of something. So we can talk about $F(x)$ defined on $(-2,2)$ with derivative $f(x)$.

Now let's look at our third function, $x(\theta) \,=\,2\sin \theta.$ We must note a few facts related to it.

• it is continuous on the interval $[-\pi / 2,\pi / 2]$.
• it is differentiable on the interval $(-\pi / 2,\pi / 2)$. (so we can have our fourth function $x'(\theta)$.)
• $x'(\theta)$ is strictly greater than zero inside its domain $(-\pi / 2,\pi / 2)$.
• we have a neat matchup of endpoints: $x(-\pi / 2) = -2, x(\pi / 2) = 2$.

All this indicates that $x(\theta)$ has an inverse function $\theta(x)$. Or in other words, the $x$'s are in one-to-one correspondence with the $\theta$'s. The upshot is, that although writing the following is the obvious choice

$$\int{f(x) dx } \,=\, F(x),$$

we are justified to rather take the point of view

$$\int{f(x) dx } \,=\, \Phi(\theta(x)).$$

(Here $\Phi$ performs the action: apply $F$ to the $x$ which corresponds to $\theta$.) Now perhaps you begin to see why a COV is more than just a notational thing. And, that this curious change in point of view is the key to making the substitution work. We need also the chain rule:

$$\frac{d}{d \theta} \Phi(\theta) \,\,=\,\, \frac{d}{d \theta} F(x(\theta)) \,\,=\,\, f(x(\theta))\,x'(\theta).$$

The third expression is a derivative in $(-\pi / 2,\pi / 2)$. Hence its integral is $\Phi(\theta)$. As your wiki page shows, $\Phi(\theta) = \theta.$

Lastly we say goodbye to $\theta$ by noting that $\theta(x) \,=\, \arcsin(x/2)$ and again that $\Phi(\theta(x)) = F(x)$.

I tried to do all this while avoiding any interpretation of 'dx'. Of course it is fine to think of $dx$ as the width of very skinny rectangles. But in the COV/substitution, you must think of the two intervals $-2 < x < 2$, $-\pi/2 < \theta < \pi/2$ and some differentiable bijection between them.

Answer 2

That is because, if you interpret $\mathrm dx$ as a small variation $\Delta x$ of $x$, and if $x$ is a function $f(\theta)$, where $f$ is a differentiable function, $f'(\theta)\Delta \theta\:$ is the best linear approximation of $\Delta x$, in the very precise sense given by asymptotic analysis that $$\Delta x=f'(\theta)\,\Delta \theta+o\bigl((\Delta\theta)^2\bigr).$$ Therefore one sets the differential element as $$\mathrm dx=f'(\theta)\,\mathrm d\theta.$$