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Thread: Integral sin(3x pi/a)sin(2x pi/a)cos(x pi/a)

  1. #1
    Senior Member Educated's Avatar
    Aug 2010
    New Zealand

    Integral sin(3x pi/a)sin(2x pi/a)cos(x pi/a)

    I'm doing a bit of quantum mechanics and I'm doing a question on the infinite square well.

    Anyway, I'm a bit stuck on this integral:

    $\displaystyle \displaystyle \sqrt{\dfrac{2}{a}} \dfrac{2}{\sqrt{a}} \int_0^a \sin \left( \dfrac{3 \pi}{a} x\right) \sin \left( \dfrac{2 \pi}{a} x\right) \cos \left( \dfrac{\pi}{a} x\right)dx$

    Does anyone know how to integrate this?

    I tried reversing the product rule for 3 functions and got this: $\displaystyle \displaystyle \int u v w' = u v w - \int u' v w - \int u v' w $
    But I didn't really get anywhere.

    I know that these functions are orthogonal to each other (except for the cosine) and so if they were all sines I can use the Kronecker Delta function to evaluate them faster, but that cosine really messes things up.

    Using Wolfram Alpha I got $\displaystyle \dfrac{\sqrt{2}}{2}$ but I want to know how to do this.
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  2. #2
    Super Member
    Jul 2012

    Re: Integral sin(3x pi/a)sin(2x pi/a)cos(x pi/a)

    remember the sum and product formula for trig functions.
    2 sinAsinB = cos(A-B) - cos ( A+B)
    2 cosAcosB = cos(A+B) + cos ( A-B)
    2 sinAcosB = sin(A+B) + sin ( A-B)
    2 cosAcosB = sin(A+B) - sin ( A-B)
    Apply the appropriate one and you will have your integral reducing to simple integral of trig ratios
    Thanks from Educated
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