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Evaluate the definite integral and it exists

  1. May 8, 2007 #1
    a=0, b= pie/3 sin theta/ cos^2 theta

    i let u= cos theta
    du/dx= -sin theta
    -du = sin theta

    sin theta/u^2 theta

    then i anti differentiated it

    -cos theta/ (1/3)(cos)^3 theta
    this is where i got stuck
     
  2. jcsd
  3. May 8, 2007 #2
    i know how to do this problem but need help with the antidifferentiation
     
  4. May 8, 2007 #3

    Hurkyl

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    Did you mean
    [tex]\int_{0}^{\pi / 3} \frac{ \sin \theta }{\cos^2 \theta}[/tex]
    ? If so, that makes no sense: there is a very important part of this notation that has been omitted.

    It looks like you were doing a substitution -- is there any particular reason why you didn't convert the expression entirely into a function of u?
     
  5. May 8, 2007 #4
    yes that is what i meant, the reason is because sin and cos confuse me
     
  6. May 8, 2007 #5
    I think I had this same problem not too long ago. sin(u)/cos(u) is a trig property that equals tan(u), and 1/cos(u) is the definition of sec(u). Therefore, the problem condenses to tan(u)sec(u)du, which you should recognize if you're on up to integration in your studies.
     
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