Does U-Substitution Function as the Inverse of the Chain Rule?

[NoahsArk]

TL;DR

I don't get it.

Can someone please give as simple an example as possible to show what U substitution is about? I know basic integration rules but don't understand the point of u-substitution. I've read that it's used to "undo the chain rule", but I don't see how, and don't see how we can spot when we'd need to reverse the chain rule for integration as opposed to just following the normal rule for integration ( adding one to the exponent and dividing by the exponent plus 1). Thanks

 

[PeroK]

TL;DR Summary: I don't get it.

Can someone please give as simple an example as possible to show what U substitution is about? I know basic integration rules but don't understand the point of u-substitution. I've read that it's used to "undo the chain rule", but I don't see how, and don't see how we can spot when we'd need to reverse the chain rule for integration as opposed to just following the normal rule for integration ( adding one to the exponent and dividing by the exponent plus 1). Thanks

I thought we went through all this in a thread some time ago?

 

[NoahsArk]

@PeroK I remember asking about the chain rule and implicit differentiation, but not about u substitution. If I did ask about it and you have a link to the prior discussion, please paste it. Thanks

 

[PeroK]

@PeroK I remember asking about the chain rule and implicit differentiation, but not about u substitution. If I did ask about it and you have a link to the prior discussion, please paste it. Thanks

Apologies, I wrongly thought you were involved in this thread.

https://www.physicsforums.com/threads/question-about-change-of-variables.1060606/#post-7065893

 

[NoahsArk]

@PeroK NP

 

[Mark44]

TL;DR Summary: I don't get it.

Can someone please give as simple an example as possible to show what U substitution is about?

Here's an example that is similar to a question that was recently asked.
$$\int x(x^2 + 3)^3 ~dx$$
You could of course multiply out the square term to get a 7th degree polynomial, but using ordinary substitution is more efficient.

Using the substitution $u = x^2 + 3$ from which we get $du = 2xdx$, we can rewrite the integral as $\int u^3 \frac 1 2 du = \frac 1 2 \int u^3 du$. You can then use the power rule for integrals to evaluate the revised integral. Once the integration is performed, undo the substitution.

 

[phyzguy]

Another example is:
$$ I = \int \sin{x} cos{x} dx $$
It's not obvious how to do this. But if I set $$ u = \sin{x}, du = \cos{x} dx $$, then we have:
$$ I = \int u du = 1/2 u^2 = 1/2 \sin^2{x} $$

 

[Mark44]

then we have: $$ I = \int u du = 1/2 u^2 = 1/2 \sin^2{x} $$

Although the above is correct, some might (incorrectly) interpret the last two expressions as $\frac 1 {2u^2}$ and $\frac 1 {2\sin^2(x)}$.

These are less prone to misinterpretation as $\frac {u^2}2$ and $\frac{\sin^2(x)}2$ or $\frac 1 2 u^2$ and $\frac 1 2 \sin^2(x)$.

 

[NoahsArk]

Thank you for the responses.

 

[phyzguy]

Elaborating a little more on my example, and how u substitution is basically the inverse of the chain rule. If you have:$$ \frac{d}{dx} \frac{sin^2(x)}{2}$$
Applying the chain rule, you get:
$$ \frac{d}{dx} \frac{sin^2(x)}{2} = \frac{d}{d(sin(x))}\frac{sin^2(x)}{2} \frac{d}{dx} sin(x) = sin(x) cos(x)$$
Do you see how this is really just the inverse of the u substitution we did above?

 

[PeroK]

Elaborating a little more on my example, and how u substitution is basically the inverse of the chain rule. If you have:$$ \frac{d}{dx} \frac{sin^2(x)}{2}$$
Applying the chain rule, you get:
$$ \frac{d}{dx} \frac{sin^2(x)}{2} = \frac{d}{d(sin(x))}\frac{sin^2(x)}{2} \frac{d}{dx} sin(x) = sin(x) cos(x)$$
Do you see how this is really just the inverse of the u substitution we did above?

The change of variables is really the inverse of the chain rule. Formally, if $f$ and $u$ are suitable functions:
$$\int_a^bf(u(x))u'(x) \ dx = \int_{u(a)}^{u(b)}f(t) \ dt$$The proof is relatively simple:

Let $F(y) = \int_0^y f(t)dt$. By the FTC (Fundamental Theorem of Calculus), $F'(y) = f(y)$.

Let $g(x) = F(u(x))$. By the chain rule: $g'(x) = F'(u(x))u'(x) = f(u(x))u'(x)$.

Hence:
$$\int_a^bf(u(x))u'(x) \ dx = \int_a^bg'(x) \ dx = g(b) - g(a) = F(u(b)) - F(u(a))$$$$ = \int_{0}^{u(b)}f(t) \ dt - \int_{0}^{u(a)}f(t) \ dt = \int_{u(a)}^{u(b)}f(t) \ dt$$