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Difficult integral for Trig Substitution

  1. Mar 20, 2009 #1
    ok i have been studying the in-depth processes of trigonometric substitution with integrals and this problem has me frusterated.

    [tex]\int x^2\sqrt{(x^2-4)} dx[/tex]

    The evaluation is clear (from an old Table of Integrals I found), but the derivation is not at all clear, which is what i want to know.

    I also tried to solve this by integration by parts, but every approach ended with an even more complicated integral, so trig substitution is probably the best choice.

    Can anyone help?
  2. jcsd
  3. Mar 20, 2009 #2
    Try the substitution u = 2sec(theta)
  4. Mar 21, 2009 #3


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    [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] so [itex]sin^2(\theta)= 1- cos^2(\theta)[/itex] and, dividing on both sides by [itex]cos^2(\theta)[/itex], [itex]tan^2(\theta)= sec^2(\theta)- 1[/itex]. The substitution [itex]x= 2sec(\theta)[/itex], as bigred09 suggested, will reduce that squareroot to [itex]2 tan(\theta)[/itex].

  5. Mar 21, 2009 #4


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    Alternatively, use the hyperbolic substitution [tex]x=2Cosh(u)[/tex]
  6. Mar 22, 2009 #5
    ok well with trig substitution, i get

    [tex]\int tan^2\theta sec^3\theta d\theta[/tex]

    which doesn't help me. Can somone solve this integral then?
  7. Mar 22, 2009 #6
    Wrong! Look what Halls said, post #3.
  8. Mar 22, 2009 #7

    Edit: Disregard this!
    Last edited: Mar 23, 2009
  9. Mar 23, 2009 #8
    @ sutupidmath:

    You're forgetting about dx/d(theta)
  10. Mar 23, 2009 #9
  11. Mar 27, 2009 #10
    right i actually forgot the coefficient 8 but that doesn't mess with the integral. and [tex]dx=sec\theta tan\theta[/tex]

    so what halls said wass valid. all i did was simplify it more. even more so it looks like this:

    [tex]8\int \frac{cos^5\theta}{sin^3\theta} d\theta[/tex]

    so uh...seriously...any ideas on solving this?
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