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Calc II integration by trig sub

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the integral from 31/2 to 3 of 1/(x) ((x2 +9))1/2


    I tried to substitute x= 3tanu, and I get down eventually to a solution that is (sec(pi/4) - sec(pi/6))

    I realize integrals can have many forms, but the form in the back of my book isnt even close to sec(arctan(3/3)-sec(arctan(31/2/3). My answer doesnt match the books. Maybe its a domain problem?

    The books answer is: 1/6ln((3-2*21/2)(7+4*31/2

    Sorry, I'm exhausted, and I do not have the energy to type out how I got from the tangent substitution to my final value.

    Any help would be great.
     
  2. jcsd
  3. Sep 11, 2009 #2

    Mark44

    Staff: Mentor

    Is this your integral?
    [tex]\int_{\sqrt{3}}^3 \frac{\sqrt{x^2 + 9}}{x}dx[/tex]
    Your substitution looks to be the correct one, but if you're not getting the same answer as in your text, possibly you made a mistake after doing the substitution. It's also possible, but less probable, that the book's answer is wrong.
     
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