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A integral consists of sin(n+1/2) n=1,2,3

  1. Aug 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi all, i encounter this integral while i am trying to find the fourier series of f(x)=ln(2*sin(x/2))


    pi sin[(n+1/2)x] sin[(n-1/2)x]
    ∫ ------------- + ------------- dx n=1,2,3......
    0 sin[(1/2)x] sin[(1/2)x]


    2. Relevant equations

    sin(A+B)=sinAcosB+sinBcosA
    sin(2A)=2sinAcosA

    3. The attempt at a solution

    the answer seems to be 2*pi, but i still don't know how to work it out..
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 29, 2009 #2
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