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Trig Substitution (?) Integral

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    j8m7mg.jpg

    The answer is:

    v2se1u.jpg


    3. The attempt at a solution


    I tried trig substitution, letting x =[tex]\sqrt{2}[/tex]tan([tex]\theta[/tex]) and using the identity 1+tan[tex]^{2}[/tex]=sec[tex]^{2}[/tex]([tex]\theta[/tex]), but couldn't get to the answer.

    Thanks for the help.
     
  2. jcsd
  3. Feb 3, 2010 #2

    Mark44

    Staff: Mentor

    Show us what you did, and we can set you straight. That looks like the right substitution.
     
  4. Feb 3, 2010 #3
  5. Feb 3, 2010 #4

    Mark44

    Staff: Mentor

    Now undo your first substitution, which was x = sqrt(2)tan(theta). It will be helpful to draw a right triangle (I didn't see one in your work). The acute angle is theta. The opp. side is x, the adj. side is sqrt(2), and the hypotenuse is sqrt(x^2 + 2). Convert your secant terms back to terms involving x, and see if that gets you to your answer.

    Your work looks pretty good - I didn't see anything obviously wrong, but I just scanned it quickly, so might have missed something. When you get your final answer, check it by differentiating it - you should get 11x^3/sqrt(x^2 + 2).
     
  6. Feb 3, 2010 #5
  7. Feb 4, 2010 #6

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
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    You forgot to cube the root 2 in the denominator of the second term in the square brackets.

    You're also off by a sign. You flipped the sign when you went from tan to sec when integrating.
     
  8. Feb 4, 2010 #7
    I got it now, thanks a lot.
     
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