Integration question - use of substitution

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Homework Help Overview

The discussion revolves around the integration of the function \(\int \frac{dx}{\sqrt{1-e^{2x}}}\), with participants exploring substitution methods and questioning the validity of their approaches.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts a substitution method but expresses concern about the correctness of their first substitution. Other participants discuss the potential for alternative substitutions and the conditions under which the integral can be evaluated.

Discussion Status

Participants are actively engaging with the problem, offering feedback on the original poster's approach and discussing the flexibility of substitution methods. There is an ongoing exploration of the requirements for valid substitutions and the implications of the integral's domain.

Contextual Notes

One participant notes that the original poster has not provided the intervals for integration, which may affect the approach taken. There is also mention of constraints related to the function's domain, specifically where \(e^{2x} < 1\).

twoflower
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Hi,

I have to find this one:

[tex] \int \frac{dx}{\sqrt{1-e^{2x}}}[/tex]

Is this right approach?

[tex] \int \frac{dx}{\sqrt{1-e^{2x}}} = \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}[/tex]

Substitution:
[tex] t = \sqrt{1-e^{2x}}[/tex]

[tex] dt = - \frac{e^{2x}}{\sqrt{1-e^{2x}}} dx[/tex]

[tex] e^{2x} = 1 - t^2\\[/tex]

[tex] \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}} = - \int \frac{dt}{1-t^2} = - \int \frac{1-t}{1-t^2}dt - \int \frac{t}{1-t^2}dt [/tex]

Substitution:
[tex] y = 1 - t^2[/tex]

[tex] dy = -2t dt[/tex]


[tex] z = 1 + t[/tex]

[tex] dz = dt[/tex]

[tex] ... = - \int \frac{dz}{z} - \frac{1}{2} \int \frac{dy}{y} = - \ln \left(1 + \sqrt{1-e^{2x}} \right) - \frac{1}{2} \ln \left(e^{2x} \right) + C[/tex]

I'm afraid that the first substitution is not ok, but could someone please give me more detailed answer?

Thank you.
 
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Well,the last sign should be a "+" in the term with "ln of e^{2x}"...

[tex]\int \frac{dt}{1-t^{2}}[/tex] can be don also using the substitution

[tex]t=\tanh u[/tex]

Daniel.
 
Last edited:
Thank you dextercioby,

could you tell me whether the use of the first substitution in my approach is ok?
 
Btw we haven't learned hyperbolic functions.
 
I told u,everything is okay,except for the last sign.

Daniel.
 
It's okay.U could do it by simple fraction decomposition,but u said u needed to do it by substitution.

Daniel.
 
Ok, thank you. But I have one doubt about my approach anyway :wink:

The theorem about substition says we can use the substitution in case we have something like this:

[tex] \int f(g(x)) g'(x) dx[/tex]

But...I'm afraid this is not exactly my case. My integral just doesn't have this form...
 
This one:

[tex] \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}[/tex]
 
  • #10
Well,look at it this way

[tex]x\rightarrow t(x)\rightarrow \frac{e^{2x}}{e^{2x}\sqrt{1-e^{2x}}}[/tex]

and u see that f(x) is your initial function & g(x) is t(x)=sqrt(1-e^(2x)) ...

U needn't worry about the form of the functions.Just make the substitutions which would provide simpler forms for the integrals.

Daniel.
 
  • #11
Ok, so we're quite free to use substitutions if it helps us. I worried about whether my substitution meets the requirements of the theorem..
 
  • #12
Since u haven't provided the intervals on which u wish to integrate that function,then u can do pretty much everything...

For example,you function can be integrated only on the domain in which

[tex]e^{2x}<1[/tex]​

,so that would set a condition on the variable u want to use as a substitution for "x"...

Daniel.
 

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