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Math Help - Recurrent formula

  1. #1
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    Recurrent formula

    I have this kind of problem. I have to find a recurrent formula for
    I(n)=integrate (tan(x))^n)dx from n=0 to a, and then I need to fine I(0) and I(1).
    I wrote it like this
    I(n)=int ((tan(x))^(n-1)*(tan(x)))dx and then tried to integrate it partially.

    u=(tan(x))^(n-1)
    du=(n-1)*((tan(x))^(n-2))/(cos(x))^2
    v=int (tan(x)dx)=-log|cos(x)|
    But, when I put it back...
    I(n)=-tan(x)*log|cos(x)|-(n-1)*int((tan(x))^(n-2)* (log|cos(x)|/(cos(x))^2)
    I have this, and I don't know what to do with this new int.
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  2. #2
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    Re: Recurrent formula

    Quote Originally Posted by wabaffet View Post
    I have this kind of problem. I have to find a recurrent formula for
    I(n)=integrate (tan(x))^n)dx from n=0 to a, and then I need to fine I(0) and I(1).
    I(n) = \int_0^a\tan^nx\,dx

    I(0) = \int_0^adx

    I(1) = \int_0^a\tan x\,dx

    You should be able to integrate those. For n\geq2,

    I(n) = \int_0^a\tan^2x\tan^{n-2}x\,dx

    = \int_0^a\left(\sec^2x-1\right)\tan^{n-2}x\,dx

    = \int_0^a\sec^2x\tan^{n-2}x\,dx - \int_0^a\tan^{n-2}x\,dx

    = \int_0^a\sec^2x\tan^{n-2}x\,dx - I(n-2)

    Now you should be able to evaluate the integral on the left...
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  3. #3
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    Re: Recurrent formula

    I did it, it's all ok at the end. Thank you.
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