0
$\begingroup$

This is a question about trig substitution used in integrals.

Because it is difficult to solve an integral when there is radical in it, we use a trig function of theta to substitute x from the original integral.

The only thing I don't understand is why the ending symbol dx, which indicates that we are integrating the function with respect to the "x" variable, is also changed into a function of d theta and then multiplied by the integral after substitution.

I think after substitution, we can just rewrite dx into d theta.

$\endgroup$
1
  • 1
    $\begingroup$ if $\theta=f(x)$, then $d\theta=f'(x)dx$ $\endgroup$ Mar 16 at 12:28
1
$\begingroup$

$dx$ and $d\theta$ are not the same thing. For example, if $x = \sin \theta$, then $\frac{dx}{d\theta} = \cos \theta $ and so $dx = \cos {\theta} d\theta$.

This is important because if that were not true then just about any substitution would become trivial and integration would be as easy as differentiation.

$\endgroup$
4
  • $\begingroup$ " then dx/dθ=cosθ and so dx=cosθdθ." why it feels like u are treating dx/d theta as dx divided by d theta? $\endgroup$ Mar 16 at 12:54
  • $\begingroup$ That is what Newton thought he was doing, and it is an effective way of getting the right results even if someone has to pull their hair a bit to make sure it's logically airtight. I was trained in physics first, so I tend to have methods that look sloppy but work. $\endgroup$ Mar 16 at 18:20
  • 1
    $\begingroup$ @mirthspritzultyrobscurantism So there are two ways of resolving this. One is simply saying that, by the fundamental theorem of calculus, and the chain rule, we have $\int_a^b g'(x)f(g(x))\textrm{d}x=\int_{g(a)}^{g(b)} f(u)\textrm{d}u$. This is a borderline trivial and completely rigorous statement. One short-hand for this could be to write "$u=g(x)$ implies $\textrm{d}u=\frac{du}{dx}\textrm{d}x$ since that's sort of how the above look if you think all you're doing is integrating $f$. And this heuristic will always yield correct substitutions. $\endgroup$ Mar 16 at 20:17
  • 1
    $\begingroup$ Another is appealing to measure theory, stating that $\textrm{d}x$ is a measure (the ordinary Lesbegue measure) and $\textrm{d}u$ is the measure such that $\textrm{d}u([a,b])=u(b)-u(a)$. In this formulation, $\textrm{d}u=\frac{du}{dx} \textrm{d}x$ is true as a statement about measures with densities and can be proven rigorously. One should note, though, that $\frac{du}{dx}$ is a function and not a measure, so although the notation is suggestive, the result isn't literally due to "fractions cancelling out". $\endgroup$ Mar 16 at 20:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.