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Homework Help: Indefinite Integral

  1. Sep 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Sorry for the poor use of Latex, I have tried to get it to work but it seems to never come out as I would like.

    Using a trigonometric or hyperbolic substitution, evaluate the following indfe nite integral,

    ∫[itex]\frac{1}{\sqrt{(x^2-1)^5}}[/itex] dx

    2. Relevant equations

    I have got down to a point where I am stuck and was wondering which path to go down next.

    3. The attempt at a solution

    let x=cosh[itex]\phi[/itex]
    dx=sinh[itex]\phi[/itex] d[itex]\phi[/itex]

    Then (x2-1) = cosh2[itex]\phi[/itex]-1
    = sinh2[itex]\phi[/itex]

    (x2-1)1/2= sinh[itex]\phi[/itex]
    (x2-1)5/2= sinh5[itex]\phi[/itex]

    therefore ∫sinh[itex]\phi[/itex]/sinh5[itex]\phi[/itex]

    Is this right so far, Do i then split the (cosec[itex]\phi[/itex])^4 into two and do the integral then.?
  2. jcsd
  3. Sep 5, 2013 #2


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    Homework Helper

    The last integral should be ##\int (\mathrm{csch} \phi)^4 d\phi##, shouldn't it?

    There might well be an easier way to go about this with hyperbolic trig identities, but I would do it with a mainly algebraic integral.

    First convert to the exponential form. I'm going to be using y instead of ##\phi## for ease of Latex:

    You have to evaluate ##\int \frac{16}{(e^y - e^{-y})^4}dy##.

    Sub ##u = e^y##. Simplify. You'll end up with a ##u^3## in the numerator, and a ##(u^2 - 1)^4## in the denominator. There's a clever way to split this up so you can integrate by parts. And this can be repeated one more time.
  4. Sep 5, 2013 #3


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    So far it looks good (in the last line it must be [itex]\text{cosech}^4 x[/itex] of course.

    My next guess is that it might help to use
    [tex]\cosh^2 x-\sinh^2 x=1[/tex]
    in the numerator of the integrand ;-).
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