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Homework Help: Integration by substitution

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Compute the indefinite integral.

    ∫(x^2 + 1)^(-5/2) dx

    3. The attempt at a solution
    I have a hunch that I need to substitute x = tan(u) but, as always, my lack of trig skills are holding me back.
     
  2. jcsd
  3. Dec 9, 2009 #2

    Mark44

    Staff: Mentor

    Yes, that's the right substition.

    If x = tan u, dx = ?
     
  4. Dec 9, 2009 #3
    So dx = sec^2 u du?
     
  5. Dec 9, 2009 #4

    Mark44

    Staff: Mentor

    Right. And you also have the du, so good job. You'll need to substitute for the x^2 + 1 part, so that's tan^2(x) + 1, right?

    When you're done with you trig substitution, your new integral will be completely in terms of u and du, with x and dx completely gone.
     
  6. Dec 9, 2009 #5
    I plugged in the x = tan u and dx = sec^2 u du, and I got
    ∫[tan^2(u) + 1]^(-5/2) * sec^2(u) du
    ==> ∫[sec^2(u)]^(-5/2) * sec^2(u) du
    ==> ∫[sec^2(u)]^(-3/2) du
    after integration I end up with (-3/2)(tanu/secu)
    If I continue on and cancel out the cosines, I end up with (-3/2)sinu.
    I don't know how to get this back in terms of x.
     
  7. Dec 9, 2009 #6
    Wait, I think I got this.
    I didn't write out all of my steps in my last reply..so I'll go back one or two, to one I skipped.
    (-3/2)[tan(u) / √sec^2(u)]
    ==>(-3/2) [tanu / √(tan^2u + 1)]
    ==>-3x / 2√(x^2 + 1)
     
    Last edited: Dec 9, 2009
  8. Dec 9, 2009 #7

    Mark44

    Staff: Mentor

    Don't use ==> when you mean =.

    Also, don't forget your constant of integration. I didn't check your answer, but you can do that. If your antiderivative is correct, you should be able to differentiate it and get back the original integrand. IOW, d/dx(-3x/2sqrt(x^2 + 1)) should equal (x^2 + 1)^(-5/2).
     
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