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Bonus Homework Problem: Integral of csc

  1. Feb 25, 2009 #1
    [tex]\int csc dx = ?[/tex]


    I just am seeking confirmation as to whether or not I did this correctly. I apologize beforehand for any formatting errors.

    [tex]\int csc(x) dx = [/tex] [tex]\int 1/sin(x) dx[/tex]
    u = sin(x)
    du = cos(x)dx
    du / cos(x) = dx

    u2 = sin2(x)
    u2 = 1 - cos2(x)
    [tex]\sqrt{}[/tex]{u^2-1} = cos(x)
    du / [tex]\sqrt {u2-1} = dx[/tex]

    Therefore

    [tex]\int du / u\sqrt{}[/tex]{u2-1}

    And, since I cannot remember the arc property, this is as for as I was able to get.
     
    Last edited by a moderator: Feb 26, 2009
  2. jcsd
  3. Feb 25, 2009 #2

    lanedance

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    Homework Helper

    try u = ln(tan(x/2))
     
  4. Feb 26, 2009 #3

    Mark44

    Staff: Mentor


    This is an interesting approach, but you have a problem in the next line. The intermediate step is: u2 - 1 = - cos2(x).
    The right side is always <= 0, so taking its square root almost always gives you imaginary numbers.

    That's an interesting approach, but there's a problem here:
     
  5. Feb 26, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I would integrate [itex]\int dx/cos(x)[/itex] by multiplying both numerator and denominator by cos(x): [itex]\int cos(x)dx/cos^2(x)= \int cos(x)dx/(1- sin^2(x))[/itex] and then let u= sin(x).
     
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