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Math Help - polar integral

  1. #1
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    Question polar integral

    8/3 times the integral (from 0 to pi) of sin^3 theta d(theta).

    Thank you very much.
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  2. #2
    Eater of Worlds
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    \int{sin^{n}(x)}dx=\frac{-1}{n}sin^{n-1}(x)cos(x)+\frac{n-1}{n}\int{sin^{n-2}(x)}dx


    You can use the formula to check if you wish.

    \int{sin^{3}(x)}dx

    \int{sin(x)sin^{2}(x)}dx

    \int{sin(x)(1-cos^{2}(x))}dx

    Let u=cos(x), \;\ du=-sin(x)dx

    \int{(u^{2}-1)}du=\frac{u^{3}}{3}-u

    Resub:

    \frac{cos^{3}}{3}-cos(x)+C

    or by the reduction formula:

    \frac{-sin^{2}(x)cos(x)}{3}-\frac{2}{3}cos(x)+C

    Same thing.
    Last edited by galactus; August 1st 2007 at 10:56 AM.
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  3. #3
    Senior Member tukeywilliams's Avatar
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     \frac{8}{3} \int_{0}^{\pi} \sin^{3} \theta \ d \theta = \frac{8}{3} \int_{0}^{\pi} (1- \cos^{2} \theta) \sin \theta \ d \theta . Let  u = \cos \theta , then  du = - \sin \theta \ d \theta or

     -\frac{8}{3} \int_{0}^{\pi} 1-u^2 \; du = u - \frac{u^3}{3} = -\frac{8}{3} \left[\cos \theta - \frac{\cos^{3} \theta}{3} \right ] .


    What would it be from  0 to  \pi ?  \frac{32}{9} .
    Last edited by tukeywilliams; August 1st 2007 at 11:09 AM.
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  4. #4
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    Thank you very much again turkeywilliam. Yes, the answer is 32/9. Thanks!
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