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Math Help - Proving the reduction formula?

  1. #1
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    Proving the reduction formula?

    How do you go about proving that

    integral of sin(X)^(2n+1) dx evaluated on the interval [0,pi/2]= pi/2 x [(2 x 4 x 6.......x (2n-1)) / (3 x 5 x 7...........x 2n)] ?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Using integration by parts to:

    I_{2n+1}=\displaystyle\int_{0}^{\pi/2}\sin ^{2n+1} x\;dx,\quad (u=\sin^{2n} x,\;dv=\sin x dx)

    you'll obtain the recurrence relation:

    I_{2n+1}=\dfrac{2n}{2n+1}I_{2n-1}


    Fernando Revilla
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