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Math Help - Integration by Reduction help! Completly confused.

  1. #1
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    Integration by Reduction help! Completly confused.

    Alright all im stuck on this integration by reduction question;

     I_n = \int_0^\Pi Sin^(2n) \theta Show that

    I_n = \frac{2n-1}{2n} I_(n-1) and hence I_n = \frac{(2n)!}{(2^nn!)^2}

    Sorry my latex isnt so good at the moment.

    Help is appricated. Thanks!
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  2. #2
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    Quote Originally Posted by simpleas123 View Post
    Alright all im stuck on this integration by reduction question;

     I_n = \int_0^\Pi Sin^2\cdot n \theta
    I assume you mean

    I_n=\int_{0}^{\pi}\sin^{2n}\theta d\theta

    There are two ways to do this.

    First method

    Write the integral as for n \ge 1

    I_n=\int_{0}^{\pi}\sin\theta \sin^{2n-1}\theta d\theta

    Use integration by parts.

    dv=\sin(\theta)d\theta \implies v=-\cos(\theta) and u=\sin^{2n-1}(\theta) \implies du=(2n-1)\sin^{2n-2}(\theta)\cos(\theta)d\theta

    This gives

    I_n=-\cos(\theta)\sin^{2n-1}(\theta)\bigg|_{0}^{\pi}+(2n-1)\int_{0}^{\pi}\sin^{2n-2}(\theta)\cos^2(\theta)d\theta

    Use the pythagorean Identity from here and you are done

    The 2nd is to use complex variables. If you want to see it this way I can show you that as well.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    I assume you mean

    I_n=\int_{0}^{\pi}\sin^{2n}\theta d\theta

    There are two ways to do this.

    First method

    Write the integral as for n \ge 1

    I_n=\int_{0}^{\pi}\sin\theta \sin^{2n-1}\theta d\theta

    Use integration by parts.

    dv=\sin(\theta)d\theta \implies v=-\cos(\theta) and u=\sin^{2n-1}(\theta) \implies du=(2n-1)\sin^{2n-2}(\theta)\cos(\theta)d\theta

    This gives

    I_n=-\cos(\theta)\sin^{2n-1}(\theta)\bigg|_{0}^{\pi}+(2n-1)\int_{0}^{\pi}\sin^{2n-2}(\theta)\cos^2(\theta)d\theta

    Use the pythagorean Identity from here and you are done

    The 2nd is to use complex variables. If you want to see it this way I can show you that as well.

    The 1st method is very good, but once i get down to

    I_n = (2n-1)\int_0^\pi sin^2n-2 - sin^2n

    I dont know were to go from there?
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  4. #4
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    Quote Originally Posted by simpleas123 View Post
    The 1st method is very good, but once i get down to

    I_n = (2n-1)\int_0^\pi sin^2n-2 - sin^2n

    I dont know were to go from there?
    Use the linear property of ingegrals, then ask yourself 2hat is the definition of

    I_{2n} \text{ and } I_{2n-2}.

    Good luck
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  5. #5
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    Don't solve this by reduction , use oxidation
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