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Thread: trig integrals

  1. November 8th 2009, 09:40 PM #1
    abilitiesz
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    trig integrals

    I just need help on one more of this type of integral:



    Can someone help me start it up? Thank you.
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  2. November 8th 2009, 10:22 PM #2
    abilitiesz
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    I figured it out. Thanks.
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  3. November 8th 2009, 10:22 PM #3
    simplependulum
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    \frac{8}{1287}


    sub \cos(x) = t

    dt = - \sin(x) ~dx

    I = \int_0^{1} t^8 ( 1 - t^2 )^2 ~dt

    = \int_0^{1} (t^8 - 2 t^{10} + t^{12} )~dt

    = \frac{1}{9} - \frac{2}{11} + \frac{1}{13}

    = \frac{1}{1287} [ 143 - 234 + 99 ] = \frac{8}{1287}
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