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Trigonometric Substitution

  1. Jul 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Use an appropriate substitution and then a trigonometric substitution to evaluate the integral:
    [tex]\int_-^{ln(4)} \frac{e^{t}dt}{\sqrt{e^{2t}+9}}[/tex]

    2. Relevant equations
    Seems to be close to the [tex]t=atan(\theta)[/tex] model
    [tex]\int \frac{e^{t}}{\sqrt{a^{2}+x^{2}}}dt[/tex]

    3. The attempt at a solution

    I've rewritten it to reflect the relevant model:
    [tex]\int \frac{e^{t}}{\sqrt{3^{2}+(??)}}dt[/tex]

    Not sure how to rewrite the e in my attempt...
     
  2. jcsd
  3. Jul 3, 2009 #2

    Dick

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    It's probably easiest to think of this in two steps. First substitute u=e^t. What's the integral in terms of u?
     
  4. Jul 3, 2009 #3
    Thanks, that really helped...

    When do that u substitution, however, do I need to rewrite my limits?

    because once I integrate, I will be witting everything in terms of t...

    so which limits do I use in the end? the originals? or the ones from the u substitution?
     
  5. Jul 3, 2009 #4

    Cyosis

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    You will have to rewrite the integration limits yes. It is not so hard to see why, however. You know that your upper limit is [itex]t=\ln 4[/itex] and you have made the substitution [itex]u=e^t[/itex]. So what value will u become?

    If you rewrite everything in terms of t you can of course use the old limits. You are however not allowed to put the original limits below and above the integral sign of the du integral. So if you do not want to change the limits calculate the indefinite integral first in terms of t then fill in the limits.
     
  6. Jul 3, 2009 #5

    zcd

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    If you don't want to rewrite the limits, just note that [tex]e^{2t} = {(e^{t})}^2[/tex] and [tex]\frac{d}{dt} [e^{t}] = e^{t}[/tex] and go directly to trig substitution.
     
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