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Substituting for the entire integrand

  1. Jul 7, 2015 #1

    I found something surprising (at least to me) while looking at the following integral:
    [tex]\int \sqrt{\frac{e^x-1}{e^x+1}} dx[/tex]
    Wolfram Alpha suggests the following substitution as the first step:
    [tex]u = \frac{1}{e^x+1}[/tex]
    Which leads to the following integral:
    [tex]\int \frac{\sqrt{1-2u}}{(u-1)u} du[/tex]
    The next substitution is:
    [tex]s = \sqrt{1-2u}[/tex]
    From there, the solution is just a partial fraction decomposition away. Thinking that one should be able to do both substitutions in one step, I plugged [itex]u[/itex] into [itex]s[/itex] and got:
    [tex]s = \sqrt{1-\frac{2}{e^x+1}} = \sqrt{\frac{e^x-1}{e^x+1}}[/tex]
    So, this is the original integrand that was substituted here! I don't remember having learned this technique. Does this approach of substituting the whole integrand have a name?

    Thank you!

    Best regards,
  2. jcsd
  3. Jul 8, 2015 #2


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    Gold Member

    This is just another change of variables. It was done in two steps just to reduce the complexity of calculations.
  4. Jul 8, 2015 #3
    I understand this - maybe I wasn't clear enough.
    What surprises me here is that, using those substitutions, we effectively substituted for the entire original integrand.
    I don't remember having seen such a thing before, and I am wondering whether this is common and in which cases it works.
  5. Jul 8, 2015 #4


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    Any time you make a substitution, u= u(x), you have to also substitute for the derivative, dx. Here, with [itex]u= \left(\frac{e^x- 1}{e^x+ 1}\right)[/itex] we have [itex]u'= \frac{1}{2}\left(\frac{e^x- 1}{e^x+ 1}\right)^{-1/2}\left(\frac{e^u- 1}{e^u+ 1}\right)dx= \frac{1}{2}\left(\frac{e^x- 1}{e^x+ 1}\right)^{1/2} dx[/itex] so that [itex]2du= \left(\frac{e^x- 1}{e^x+ 1}\right)^{1/2} dx[/itex], what you are referring to as "substituting for the entire integral".

    That is, basically, a result of the fact that [itex]e^x[/itex] is its own derivative. A much simpler example would be [itex]\int e^x dx[/itex]. Making the substitution [itex]u= e^x[/itex] which give [itex]e^x dx= du[/itex] and the integral becomes [itex]\int du[/itex].
    Last edited by a moderator: Jul 8, 2015
  6. Jul 9, 2015 #5
    Thank you! It is clear to me now.
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