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Integration by Trigonometric Substitution.

  1. Oct 13, 2007 #1
    I'm not sure about answer.It looks very strange.
    1. The problem statement, all variables and given/known data

    [tex]\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}[/tex]





    3. The attempt at a solution

    for u=lnx-->u'=1/x
    [tex]\int \frac{du}{\sqrt{1+u^2}}[/tex]
    substituting [tex]u=tan\theta[/tex]

    [tex]=\int \frac{d\theta}{cos\theta}=ln|sec\theta+tan\theta|[/tex]

    [tex]\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}=ln|\sqrt{-1}|[/tex]
     
    Last edited: Oct 13, 2007
  2. jcsd
  3. Oct 13, 2007 #2

    learningphysics

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    Your formula looks right but your final results isn't. Didn't you actually use u = tan(theta) ?

    what limits do you get for theta?
     
  4. Oct 13, 2007 #3
    I just put to the
    [tex]ln|lnx+\sqrt{lnx-1}|[/tex]
     
  5. Oct 13, 2007 #4

    learningphysics

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    Did you use u = tan(theta) or u = sec(theta) ?
     
  6. Oct 13, 2007 #5
    Sorry i typed wrongly.I used u=tan(theta)
     
  7. Oct 13, 2007 #6

    learningphysics

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    ok. so this part is wrong

    [tex]ln|lnx+\sqrt{lnx-1}|[/tex]

    fix your substitution.
     
  8. Oct 13, 2007 #7

    Gib Z

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    Express sec in terms of tan again.
     
    Last edited: Oct 13, 2007
  9. Oct 13, 2007 #8
    but if I change limits
    [tex]\int_{0}^{\frac{\pi}{4}}sec\theta d\theta=ln2[/tex]
     
  10. Oct 13, 2007 #9

    Gib Z

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    I don't know what you just did, but continue as you were before, you had the right anti derivative: ln |tan O + sec O|, but you didn't replace the original variable back in properly for the sec O.
     
  11. Oct 13, 2007 #10

    dextercioby

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    Just use a "sinh" substitution when you're left only with the sqrt in the denominator and you're done.
     
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