when should a substitution of the form dx =g(u)du be used?

[Al-Layth]

I remember watching a YT video where there was an integrand ( i forgot what it was lol) that was two functions composed.
The writer let
[imath]u[/imath]= inside function, [imath]f(x)[/imath]

got the expression for
[math]dx = \frac{1}{f'(x)}du[/math]
but instead of substituting both [imath]u[/imath] and [imath]dx[/imath] into the integrand as is normally done, he instead continued by applying the inverse of the inside function to the u formula:

[math]f^{-1}(u)=f^{-1}(f(x)) \longrightarrow x[/math]
i think i have written that wrong. what I mean is the inverse of f(x) is applied to both sides, on the left u becomes the argument.

and then he subbed that into the [imath]dx[/imath] formula:

[math]dx = \frac{1}{f'(f^{-1}(u))}du[/math]
and then subbed this and the first equation, u=f(x) into the integrand to simplify it.

problem is I don't know when to use it,
what kind of integrand forms does it simplify ? thx

 

[Steven G]

I guess the integral was of the form \(\displaystyle \int g(f(x))dx\)

If u=f(x), then du = f'(x)dx or du/f'(x) = dx

The integral becomes \(\displaystyle \int \dfrac{g(u)du}{f'(x)}\)

f^-1(u) = f^-1(f(x))= x (no arrows for equal signs!!!!) This says x = f^-1(u).

Then the integral becomes \(\displaystyle \int \dfrac{g(u)du}{f'(f^{-1}(u))}\)

Now to answer your question: The answer is simple! You use it when this new integral is easier to solve than the original one (of course you will need to have a composite function). I can't think of an example but hopefully someone will come along with one.

In mathematics it is not always when will it be useful--it could be used very infrequently- but rather the fact that it is theoretically correct. That is, what you presented here is something that I enjoyed seeing. Some people like sports cars, others like paintings and some people like nice mathematical proofs.

 

[Al-Layth]

I guess the integral was of the form \(\displaystyle \int g(f(x))dx\)

If u=f(x), then du = f'(x)dx or du/f'(x) = dx

The integral becomes \(\displaystyle \int \dfrac{g(u)du}{f'(x)}\)

f^-1(u) = f^-1(f(x))= x (no arrows for equal signs!!!!) This says x = f^-1(u).

Then the integral becomes \(\displaystyle \int \dfrac{g(u)du}{f'(f^{-1}(u))}\)

Now to answer your question: The answer is simple! You use it when this new integral is easier to solve than the original one (of course you will need to have a composite function). I can't think of an example but hopefully someone will come along with one.

In mathematics it is not always when will it be useful--it could be used very infrequently- but rather the fact that it is theoretically correct. That is, what you presented here is something that I enjoyed seeing. Some people like sports cars, others like paintings and some people like nice mathematical proofs.

thx

but how can we tell this technique will work to simplify without actually trying it?

 

[topsquark]

thx

but how can we tell this technique will work to simplify without actually trying it?

Unless you get lucky, you can't. A lot of time spent on finding ways to integrate a function are spent trying various techniques. When you see a source doing the problem and they post the "correct" way to solve it either a) they've got a lot of experience or b) they tried several ways and this is one that worked.

Except for basic forms you don't see a lot of methods of integration that say, when you have ... as an integrand, then do ....

-Dan

 

[Steven G]

...without trying what??
We started with an integral in one form, made a u-substitution to write the integral into another form. What is there to try out? We proved that the forms are equivalent.

 

[Al-Layth]

...without trying what??
We started with an integral in one form, made a u-substitution to write the integral into another form. What is there to try out? We proved that the forms are equivalent.

Dude