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(cosx)^2 half angle formula

  1. Feb 7, 2009 #1
    y=sin2x bounded by x=0,x=pi,y=0 , revolved around the x axis

    cross section A=pi(sin2x)^2

    latex2png.2.php?z=100&eq=pi%5Cint_%7B0%7D%5E%7Bpi%2F2%7D%20(sin(2x))%5E2.jpg

    latex2png.2.php?z=100&eq=4pi%5Cint_%7B0%7D%5E%7Bpi%2F2%7Dsin%5E2xcos%5E2x.jpg
    taking u = sinx ; du=cosxdx

    im unclear on how to proceed in this case where du needs to satisfy (cosx)^2
    the problem hints to use a half angle formula
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 7, 2009 #2
    Re: sin2x,bound{x=0,x=pi,y=0}

    Try the identity cos(2x) = 1 - 2sin2(x). It can be rearranged to be a half-angle formula. :smile:
     
  4. Feb 7, 2009 #3
    Re: sin2x,bound{x=0,x=pi,y=0}

    still i am unclear on how to proceed here
     
  5. Feb 7, 2009 #4
    Re: sin2x,bound{x=0,x=pi,y=0}

    Try using the Pythagorean identity to get sin2(x) - sin4(x), then apply the half angle formula to get cosines (twice for the second term) that are not squared.
     
  6. Feb 7, 2009 #5

    Dick

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    Re: sin2x,bound{x=0,x=pi,y=0}

    slider142 is trying to say that since sin^2(x)=(1-cos(2x))/2, sin^(2x)=(1-cos(4x))/2. That's pretty easy to integrate.
     
  7. Feb 8, 2009 #6
    Re: sin2x,bound{x=0,x=pi,y=0}

    perfect
     
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