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Simple integral

  1. May 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Is there a simple way to evaluate

    S x^5 / [rt(x^2-ax)] dx ?

    That is, the indefinite integral of (x^5) / [square root of (x^2-ax)].

    3. The attempt at a solution

    My idea was to rewrite it as x^(9/2) / [rt(x-a)] and then do the substitution u = rt(x-a).
    Then you get

    S 2(u^2+a)^(9/2) du

    But this too is difficult to evaluate.
    Last edited: May 3, 2008
  2. jcsd
  3. May 3, 2008 #2


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    Welcome to PF!

    Hi Howsertal! Welcome to PF! :smile:

    Hint: it's always worth trying completing the square … that is, putting (x² - ax) in the form ((x-b)² - c). :smile:
  4. May 3, 2008 #3
    Hmm okay, so that puts it in the form

    ∫ x^5 / √[(x-(a/2))² - (a²/4)] dx

    One possible substitution seems to be u = x-(a/2) but that puts it in the form

    ∫(u+(a/2))^5 / [u²-(a/2)²] du = ∫ (u+(a/2))^(9/2)) / [u-(a/2)] du

    which doesnt seem alot better than what we started with. A trig substitution of the form
    [x-(a/2)] = (a/2)secψ in the first integral above would yield

    ∫ [(((a/2)secψ) + (a/2))^5)/tanψ] secψtanψ d ψ

    = ∫ (((a/2)secψ +(a/2))^5 secψ d ψ

    which doesn't seem to yield easily either =/. Those are the only obvious substitutions I can think of. Is there a trick to this or am I just not seeing something? Thanks a lot for your help (btw having the list of symbols in your sig is a great idea!)
    Last edited: May 3, 2008
  5. May 3, 2008 #4


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    secant= 1/cosine. And odd powers of trig functions are relative easy to integrate.
  6. May 3, 2008 #5


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    Hi Howsertal! :smile:
    Looks ok to me … :smile:

    Hint: what is (d/dψ)(sec^{n}ψ tanψ)?

    (yes … I tried to get them to put the symbols on the Reply page … but this is almost as good … and I only use the signature if I need it! :biggrin:)
  7. May 3, 2008 #6
    Do you mean that we can get

    ∫ (((a/2)secψ +(a/2))^5 secψ d ψ

    into an expression just involving odd powers of trig functions?
  8. May 3, 2008 #7

    That derivative came out to
    [sec^{n+2}ψ](1+n) - nsec^{n}ψ

    which seems like it could be helpful but I'm not sure how. Sorry if I seem really dense.

    Last edited: May 3, 2008
  9. May 3, 2008 #8


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    Hi Howsertal! :smile:

    Well, you want to integrate (secψ +(a/2))^5 secψ, which is powers of secψ up to the sixth, and you now know that:

    ∫((n+1)sec^{n+2}ψ - nsec^{n}ψ) dψ = sec^{n}ψ tanψ :smile:
  10. May 3, 2008 #9
    Ah i see, thank you very much. So that's a nifty way of getting expressions for all the integrals of powers of secant recursively.
    Last edited: May 3, 2008
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