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How to evaluate -arctan(cosx) from ∏/2 to ∏

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    The direction is the evaluate the integral, this isn't really a calculus issue, it's more of a trig issue.
    ∫sinx(dx)/(1+cos^2x) from ∏/2 to ∏

    2. Relevant equations
    (1/a)arctan(u/a)+c



    3. The attempt at a solution
    I did all the integrating and ended up at
    -arctan(cosx) from ∏/2 to ∏
    this is where I'm stuck, I know I'm suppose to use the fundamental theorum of Calculus but I don't know what to do once I plug in ∏/2 and ∏. How do I generate values out of this? Do I draw a triangle? Do I use the unit circle? If so, how would I use it
     
  2. jcsd
  3. Feb 28, 2012 #2

    Mark44

    Staff: Mentor

    This seems pretty straightforward.
    -arctan(cos([itex]\pi[/itex])) - (-arctan(cos([itex]\pi[/itex]/2)))

    What is cos([itex]\pi[/itex])? cos([itex]\pi[/itex]/2)?
     
  4. Feb 28, 2012 #3
    Cos ∏ is -1
    Cos ∏/2 is 0 I believe
    so then I would take the Arctan(-1)-Arctan(0)
    So I would just find on the unit circle where tangent equals -1 and 0, then subtract?
     
  5. Feb 28, 2012 #4

    Mark44

    Staff: Mentor

    Right.
     
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