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Substitution technique for integral

  1. Dec 21, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to use a substitution technique and then the table of integration to integrate the following:
    ∫ x5 √(x4 – 4) dx

    and i am given a hint which is x5 = (x3)(x2)

    I would assume that
    u = x4 – 4, then du = 4x3
    and that at some point a2 = 4 and a = 2

    However, the x5 = (x3)(x2) is confusing the heck out of me. Need help!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 21, 2009 #2


    Staff: Mentor

    What do a^2 and a have to do with anything?

    The hint looks to me like it's guiding you to use integration by parts. Is that a technique that you have learned yet? I wouldn't advise it unless you have exhausted the possibilities for ordinary substitutions, however.
  4. Dec 21, 2009 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What, so you have a limited table of integrals that doesn't include this and you are suggested a substitution to give you a form you can find in your table?

    I think the point of the x5 = x3x2 is to use the x3 as you have suggested with your u = x4 - 4. You can substitute for the extra x2 like this:

    x4 = u+4, x2 = sqrt(u+4)

    This will give you a pure u integral that may be in your table.
  5. Dec 21, 2009 #4
    The reason i used a^2 = 4 and a = 2 is because in the book they used a number of different examples where they changed the number to a letter.

    We have touched on integration by parts but everything we did learned involved using
    ex so I am not really sure how to use it here.
  6. Dec 21, 2009 #5
    It can be done with trig substitution I believe. Setting x^4/4 to (x^2/2)^2 and substituting for cos(theta).

    ∫ x5 √(x4 – 4) dx

    ∫ x5 2√(x4/4 – 1) dx

    ∫ x5 2√((x2/2)^2 – 1) dx

    d/dx cosΘ = d/dx x^2/2]
    [2cosΘ] = x^2]
    [4cos^2(Θ)] = x^4]

    -sinΘ dΘ = 2x dx

    ∫ x5 2√((x2/2)^2 – 1) dx

    ∫ x^4(2x)√((x2/2)^2 – 1) dx
    ∫4cos^2(Θ) (-sinΘ)sinΘ dΘ

    And then from there it's pretty simple... Is that right?
    Last edited: Dec 21, 2009
  7. Dec 21, 2009 #6
    I haven't learnt how to use trig substitutions yet so that will not help. Thanks for tip anyways.
  8. Dec 21, 2009 #7
    x5√(x4 - 4) = x2∙x3√(x4 - 4)
    and you can use integration by parts on that.
  9. Dec 22, 2009 #8
    Would this be correct?
    ∫ x5 √(x4 – 4) dx
    = ∫ ((x2*x2*x √(x4 – 4)) dx

    Let u = x2, then du/2 = xdx
    = ∫ 1/2 (u2√u2 – 4) du

    Let a = 4 and a2 = 2
    = ∫(u2√u2 – a2) du
    = [x(u2 – a2)3/2 / 4] + [a2x√(u2 – a2) / 8] – (a4/8) ln(x + √(u2 – a2))
  10. Dec 22, 2009 #9


    Staff: Mentor

    It looks like you are using formula #10 here.

    If so, you final result should not have u or a in it. You also need the constant of integration, which is needed in all indefinite integrals.
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