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Alternative Derivation of sin integral

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi guys; I'm just dealing with fourier series and they evaluate integrals such as ∫sin(nπx/L)dx from 0 to L as (L/nπ)[1-(-1)^n]. Can someone please tell me how to get to this conclusion or point me in the direction of a resource that will show me? Additionally I need to solve similar integrals like ∫sin^2(nπx/L)dx and ∫xsin(nπx/L) so I really need to know the technique. Thanks in advance.

    2. Relevant equations
    ∫sin(nπx/L)dx from 0 to L = (L/nπ)[1-(-1)^n]

    3. The attempt at a solution
    I would traditionally evaluate ∫sin(nπx/L)dx from 0 to L as [L-Lcos(nπ)]/nπ
     
  2. jcsd
  3. Nov 9, 2015 #2

    LCKurtz

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    So you have ##\frac{L}{n\pi}(1-\cos(n\pi))##. How does ##\cos(n\pi)## compare with ##(-1)^n##?
     
  4. Nov 9, 2015 #3

    SammyS

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    If n is even, then cos(nπ) = 1.

    If n is odd, then cos(nπ) = -1.
     
  5. Nov 10, 2015 #4
    I see now! Thank you LCKurtz and SammyS, I appreciate the help.
     
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