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Math Help - recursive formula

  1. #1
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    recursive formula

    could someone show me how to derive the recursive formula for
    (cos x )^n?


    thank you!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    could someone show me how to derive the recursive formula for
    (cos x )^n?


    thank you!
    I assume you mean \int\cos^n(x)\text{ }dx. Try using integration by parts.
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  3. #3
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    yes.

    thats what i did. and i kept getting stuck as i was not able to convert the formula to a decreasing function. i kept getting (cos x )^ (n+1) and using 1- sin^2 x = cos^2 x doesnt help..

    this is my working:

    integrate cos(x) (cos x)^(n-1)
    = sin(x) (cos x)^(n-1) +(n-1) integrate sin^2 x (cos x)^(n-1) dx
    =sin(x) (cos x)^(n-1) +(n-1) [ - cos ^n x sin x (1/n) + integrate (cos x)^(n+1) (1/n) dx ]

    the answer that i'm supoosed to get is

    1/n ( cos x)^(n-1) sin x + ((n-1)/n) integrate (cos x )^(n-2) dx

    but i cant seem to get that no matter how i try..
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    I assume you mean \int\cos^n(x)\text{ }dx. Try using integration by parts.
    \int{\cos^n{x}\,dx} = \int{\cos^2{x}\cos^{n - 2}{x}\,dx}

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

     = \int{\cos^{n - 2}{x} - \sin^2{x}\cos^{n - 2}{x}\,dx}

     = \int{\cos^{n - 2}{x}\,dx} - \int{\sin^2{x}\cos^{n - 2}{x}\,dx}

    Use integration by parts on the second integral, letting

    u = \sin{x} and dv = \sin{x}\cos^{n - 2}{x}.
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