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Homework Help: Integration question - use of substitution

  1. Mar 27, 2005 #1
    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.
     
  2. jcsd
  3. Mar 27, 2005 #2

    dextercioby

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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: Mar 27, 2005
  4. Mar 27, 2005 #3
    Thank you dextercioby,

    could you tell me whether the use of the first substitution in my approach is ok?
     
  5. Mar 27, 2005 #4
    Btw we haven't learned hyperbolic functions.
     
  6. Mar 27, 2005 #5

    dextercioby

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    I told u,everything is okay,except for the last sign.

    Daniel.
     
  7. Mar 27, 2005 #6

    dextercioby

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

    Daniel.
     
  8. Mar 27, 2005 #7
    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...
     
  9. Mar 27, 2005 #8

    dextercioby

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    Which integral...?

    Daniel.
     
  10. Mar 27, 2005 #9
    This one:

    [tex]
    \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}
    [/tex]
     
  11. Mar 27, 2005 #10

    dextercioby

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    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.
     
  12. Mar 27, 2005 #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..
     
  13. Mar 27, 2005 #12

    dextercioby

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    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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