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Tricky Sine integral

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Any ideas for how to solve the following integral?

    $$\int_{0}^{\pi}\sin{n x}\sin{x}^3 dx$$

    where n is a positive integer

    2. Relevant equations

    Various sine and cosine identities

    3. The attempt at a solution

    I haven't much of a clue how to solve the integral. Its an odd function times an odd function which gives an even function, over a symmetric range (at least symmetric for the Sine function or perhaps portions of the function's n-values).

    I tried clearing out a sin^2 from the integral by using the double-angle formula. It didn't really break down into anything that lead to any obvious results for the integrals.

    Thanks ahead of time.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 7, 2013 #2

    vanhees71

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    2016 Award

    I guess that this is a typo, because I don't think that you can find a closed solution for that integral. So I think you want to solve
    [tex]I=\int_0^{\pi} \mathrm{d} x \; \sin(n x) (\sin x)^3.[/tex]
    In this case it may be helpful to use
    [tex](\sin x)^3=\frac{1}{4}[3\sin x - \sin(3 x)].[/tex]
     
  4. Jul 7, 2013 #3
    Yes that was a typo, sorry about that.

    Cheers, I think I have done something like that already. I used the exponential forms for the sine functions and factored them all together. I am pretty sure that this gives that identity. In fact, it is very similar.

    Once I factored together the terms, I found that only for the values of n=1 and n=3 does the integral have any value at all. Which are precisely the coefficients inside of the identity.

    Though I am not completely satisfied with this answer. Because now it requires going back to the original and solving the integral for those specific values.
     
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