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Techniques of inegration,

  1. Nov 15, 2005 #1
    techniques of inegration, plz help

    in the attachment there is a problem related to techniques of integration, i fully understand everything until near the end of the problem they answer says

    (1/6)o - (1/12)sin2o + C = (1/6)o - (1/6)sino coso + C

    and then they convert the sinocoso to sec-1

    this part is really confusing...can you help me plz...

    thank you
  2. jcsd
  3. Nov 16, 2005 #2


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    But... there's no attachment? :confused:
  4. Nov 16, 2005 #3


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    Do you know the "double angle formula":
    [tex]sin(2\theta)= 2sin(\theta)cos(\theta)[/tex]

    That was what was used.

    I'm not clear what you mean by "convert the sin([itex]\theta[/itex])cos([itex]\theta[/itex]) to sec-1"
    Of course, sec([itex]\theta[/itex]) is defined as [itex]\frac{1}{cos(\theta)}[/itex] so if that "-1" is meant as reciprocal rather than "inverse function", it is true that
    [tex]sin(\theta)cos(\theta)= \frac{sin(\theta)}{sec(\theta)}[/tex]
  5. Nov 16, 2005 #4


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