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

  • #1
NoahsArk
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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
 
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  • #2
NoahsArk said:
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?
 
  • #3
@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
 
  • #5
  • #6
NoahsArk said:
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.
 
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  • #7
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:
$$ I = \int u du = 1/2 u^2 = 1/2 \sin^2{x} $$
 
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  • #8
phyzguy said:
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)##.
 
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  • #9
Thank you for the responses.
 
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  • #10
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?
 
  • #11
phyzguy said:
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:
$$\int_a^bf(u(x))u'(x) \ dx = \int_a^bg'(x) \ dx = g(b) - g(a) = F(u(b)) - F(u(a))$$$$ = \int_{0}^{u(b)}f(t) \ dt - \int_{0}^{u(a)}f(t) \ dt = \int_{u(a)}^{u(b)}f(t) \ dt$$
 
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