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Trig substitution integral (I hope)

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data

    I got to a place in a problem where I need to do a sticky integral, and I'm hoping I can use a trig substitution. If not, I will need to solve the main problem another way :(

    [tex] \int_0^\infty \sqrt{1+(e^{-\theta })^2} \; \mathrm{d} \theta [/tex]

    2. Relevant equations

    [tex] 1+\tan ^2 \theta =\sec ^2 \theta [/tex]

    3. The attempt at a solution

    Can I let [itex] e^{- \theta } = \tan \phi [/itex] ?
    if so, does [itex] \mathrm{d} \theta = \sec ^2 \phi \; \mathrm{d} \phi [/itex] ?

    And then, do I have
    [tex] \int \sec ^3 \phi \; \mathrm{d} \phi [/tex] ?
     
  2. jcsd
  3. Jan 18, 2012 #2

    Dick

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    Aside from the fact you can tell the integral is divergent just by looking at it, d(exp(-t))=(-exp(-t)*dt).
     
  4. Jan 18, 2012 #3

    I like Serena

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    Any particular reason you want to do a trig substitution?

    Myself, I'm not a fan of trig substitutions.
    Usually my first step is take a part of the expression and call it "u".
    Do a substitution and see what you are left with.
    I see a trig substitution more as a last resort.
     
  5. Jan 18, 2012 #4
    Maybe I should start at the beginning.

    I need to find the total length of the spiral [itex] r(\theta )=e^{-\theta } [/itex] for [itex] \theta \in [0,\infty ) [/itex]

    There is a formula for arc length, but I don't necessarily have to use it if there is another way:

    [tex] L=\int \sqrt{1+(r')^2} \; \mathrm{d} \theta [/tex]

    I'm concerned right away about the infinity.
     
    Last edited: Jan 18, 2012
  6. Jan 18, 2012 #5

    I like Serena

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    Hmm, that is not the right formula for the length of your curve.
    I suspect you are mixing up the formula with a cartesian version.
     
  7. Jan 18, 2012 #6
    Alas. So, how about [tex] L=\int_0^\infty \sqrt{r^2+(\frac{dr}{d\theta} )^2} \; \mathrm{d} \theta [/tex]

    in which case I will be computing
    [tex] \int_0^\infty \sqrt{ e^{-2 \theta } + (-e^{-\theta } )^2 } \; \mathrm{d} \theta [/tex] ?
     
  8. Jan 18, 2012 #7

    I like Serena

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    That looks much better! ;)
     
  9. Jan 18, 2012 #8
    Okay, thanks. night-night! :smile:
     
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