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Integral trouble trig

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Integral of dt/ (√(t^2 -6t + 13)

    2. Relevant equations
    I sub
    v = t-3, and v = 2tan(θ)




    3. The attempt at a solution

    first I completed the square of t^2 -6t + 13, I got (t-3)^2 +4

    Also I say v = t-3
    ∫ dv/ (√(v^2 +4)
    I then sub in trig

    ∫(2sec^2(θ)dθ / 2√(tan^2(θ) +1) = ∫ (sec^2(θ)/ secθ)dθ

    = ln|secθ +tanθ| = ln|(√v^2+4)/2 + v/2| edit sec here
    = ln| √((t-3)^2 +4/2) + (t-3)/2| +c
    I feel it isn't correct. What did I do?
     
    Last edited: Sep 27, 2013
  2. jcsd
  3. Sep 27, 2013 #2

    vela

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    One thing you did was write ##\sec\theta## in terms of ##v## incorrectly.
     
  4. Sep 27, 2013 #3
    OK, I fixed it. I flipped it back it was OK on my paper just not when I rewrote it on the forum.
     
  5. Sep 27, 2013 #4

    vela

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    Try differentiating your result and seeing if you recover the integrand. That's the easiest way to check your answer.
     
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