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Integeal of Sin^2

  1. Mar 28, 2012 #1
    Hi guys,


    I have a problem with the following integrals( the integral is between [-L,L]) :

    1) ∫sin(( nπ/L )x).sin(( mπ/L )x) dx = L when m = n

    and :

    2) ∫sin(( nπ/L )x).sin(( mπ/L )x) dx= 0 when m ≠ n

    I know that in the first equation we have :

    ∫sin^2(ax) dx = (x/2) - (sin2ax/4a)

    but it doesn't work like that, could you please help me with it?

    and for the second equation I think when m≠n then m or n should be even and one of the sin will be 0 so the integral will be 0, as in second equation. Am I right about it?

    Thanks in advance.
    Sophia
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 28, 2012 #2
    What do you mean it doesn't work like that? Isn't that completely correct?

    It should work for any m and n -- perhaps integrating in parts might help.
     
  4. Mar 28, 2012 #3

    epenguin

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    Gold Member

    Yes, i was going to say, it does work like that.

    This exercise is meant to introduce you to a very major mathematical theme.

    Apart from calculations, for the case m ≠ n do a sketch of sin(( nπ/L )x) and sin(( mπ/L )x) on same fig. over nm cycles*. e.g. n = 1, m = 2 or n = 2, m = 3. Look at any value of sin(( nπ/L )x) - there are general several. Look what the value of sin(( mπ/L )x) is at that x. For every value some you will find that at some of your x points sin(( mπ/L )x) is exactly minus what it is at other such points. If you think about it you may realise this has to be. For every point. Result: this product, and hence your definite integral averages out as 0. On the other hand when n = m you have a square which is never negative so the integral is not 0 in this one case.

    Very useful conclusion you will hear no end of later.


    *or lcm of n, m
     
    Last edited: Mar 28, 2012
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