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I need help, trignometric substitutions problem

  1. Aug 24, 2008 #1
    1. The problem statement, all variables and given/known data
    ∫1/ ((t^3) sqrt t^2-1) * dt


    2. Relevant equations
    sqrt x^2 - a^2 , x=a sec theta , substitution 0 less than or equal to theta less than pi/2
    identity, sec^2 theta-1 = tan^2 theta


    3. The attempt at a solution
    ok here we go
    t= sec theta , dt= sec theta * tan theta
    sqrt t^2-1= sqrt sec^2 theta -1 = tan theta

    so i plug it in now
    ∫1/ (t^3) sqrt t^2-1 * dt = ∫dt/ (t^3) sqrt t^2-1=∫ sec theta * tan theta/ sec^3 theta * tan theta

    =∫(1/sec^2 theta) * d theta

    what do i do now

    do i put in 1+ tan^2x for sec^2 theta
    and use the u substitution
     
  2. jcsd
  3. Aug 24, 2008 #2

    statdad

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    If you have

    [tex]
    \int \frac{1}{\sec^2 \theta} \, d\theta}
    [/tex]

    what do you know about the secant function in terms of something other than tangent?
     
  4. Aug 24, 2008 #3
    do i turn it into cos^2 theta
     
  5. Aug 24, 2008 #4

    statdad

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    Yes -

    [tex]
    \int \frac 1 {\sec^2 \theta} \, d\theta = \int \cos^2 \theta \, d\theta
    [/tex]

    What can be done with [tex] \cos^2 \theta [/tex]?
     
  6. Aug 24, 2008 #5
    it'll equal 1/2 (1+2cos) theta right
     
  7. Aug 24, 2008 #6

    statdad

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    I'm having a difficult time parsing your final comment, but if you wrote

    [tex]
    \cos^2 \theta = \frac 1 2 \left( 1 + 2 \cos \theta \right)
    [/tex]

    I would advise you to check the double-angle formula again (check what the
    statement above says for [tex] \theta = 0 [/tex] if you don't see why it is incorrect.) If I misinterpreted your work I apologize.
     
  8. Aug 24, 2008 #7
    That's what I meant. It's ok apology accepted.
     
  9. Aug 24, 2008 #8

    tiny-tim

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    erm … statdad was tactfully apologising for suggesting that you'd got it wrong … which you have …

    [tex]\cos^2 \theta = \frac 1 2 \left( 1 + 2 \cos \theta \right)[/tex] is wrong.

    Try again! :smile:
     
  10. Aug 24, 2008 #9

    statdad

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    As TinyTim points out (thanks, by the way :smile:)

    [tex]
    \cos^2 \theta = \frac 1 2 \left( 1 + 2 \cos \theta \right)
    [/tex]

    is not correct .
    Look very carefully at your reference book and compare the version in it to the one you wrote and we've reposted.
     
  11. Aug 24, 2008 #10
    Ok thank you.
     
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