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Math Help - stuck on integration

  1. #1
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    Question stuck on integration

    How do you integrate:
    sin(x)sin(nx)dx between Pi and Pi/2 when n doesnt equal 1

    by parts doesnt work does it?
    any help would be great
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  2. #2
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    How do you integrate:
    sin(x)sin(nx)dx between Pi and Pi/2 when n doesnt equal 1

    by parts doesnt work does it?
    any help would be great
    Remember that:

     \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B) (1)

     \cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B) (2)

    Now (2)-(1) gives:

     \cos(A-B) - \cos(A+B) = 2\sin(A)\sin(B)

    Hence:

    \sin(A)\sin(B) = \frac{ \cos(A-B) - \cos(A+B)}{2}

    Which means

     \sin(x)\sin(nx) =  \frac{ \cos(x - nx) - \cos(x+nx)}{2} = \frac{ \cos(x(1-n)) - \cos(x(1+n))}{2}

    So your integral becomes

     \int_{\frac{\pi}{2}}^{\pi} \frac{ \cos(x(1-n)) - \cos(x(1+n))}{2}\,dx  = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \cos(x(1-n))\,dx - \frac{1}{2}\int_{\frac{\pi}{2}}^{\pi} \cos(x(1+n)) \,dx

    Which should be fairly simple if you remember that  \int \cos(kx) \,dx = \frac{\sin(kx)}{k} + C
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  3. #3
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    How do you integrate:
    sin(x)sin(nx)dx between Pi and Pi/2 when n doesnt equal 1

    by parts doesnt work does it?
    any help would be great
    Use the fact that

    \sin nx \sin mx = \frac{1}{2} \cos (m-n) x - \frac{1}{2} \cos (m+n) x
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  4. #4
    MHF Contributor chisigma's Avatar
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    One possibility is the application of the famous formula...

    \sin \alpha \cdot \sin \beta = \frac{1}{2} \cdot \{ \cos (\alpha - \beta) - \cos (\alpha + \beta)\}

    Kind regards

    \chi \sigma
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  5. #5
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    thanks everyone
    got it now
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