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Math Help - trig integrals

  1. #1
    Junior Member
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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. #2
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    I figured it out. Thanks.
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  3. #3
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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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