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Math Help - Reduction method of integration

  1. #1
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    Reduction method of integration

    Establish a reduction formula for the following integrals.

    \int\tan^n x dx (use tan^2 x\equiv\sec^2 x -1)

    \int \sec^n x dx

    I figure that both are similar, and that if I can figure out one i should be able to do the other.
    The examples i was given would give the method
    \int\tan^n x dx=\int\tan x\tan^{n-1} x dx and then use integration by parts, but I'm not sure about this one.
    I tried
    \int\tan^n x dx=\int\tan^2 x\tan^{n-2} x dx
    =\tan^{n-2}x(\tan x -x)-(n-2)\int(\sec^2 x \tan^{n-3} x)(\tan x -x)
    But i could not work it out to the answer which is
    (n-1)(I_n+I_{n-2})=\tan^{n-1} x, I_n=\int\tan^n x dx and I_{n-2}=\int\tan^{n-2} x dx

    Thanks!
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  2. #2
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    \displaystyle{\int{\tan^n{x}\,dx} = \int{\tan^2{x}\tan^{n-2}{x}\,dx}}

    \displaystyle{= \int{(\sec^2{x} - 1)\tan^{n - 2}{x}\,dx}}

    \displaystyle{= \int{\sec^2{x}\tan^{n - 2}{x} - \tan^{n - 2}{x}\,dx}}

    \displaystyle{= \int{\tan^{n - 2}{x}\sec^2{x}\,dx} - \int{\tan^{n - 2}{x}\,dx}}

    \displaystyle{= \int{u^{n - 2}\,du} - \int{\tan^{n - 2}{x}\,dx}} after making the substitution u = \tan{x}

    \displaystyle{= \frac{u^{n - 1}}{n - 1} - \int{\tan^{n - 2}{x}\,dx}}

    \displaystyle{ = \frac{\tan^{n - 1}{x}}{n - 1} - \int{\tan^{n - 2}{x}\,dx}}.


    So \displaystyle{\int{\tan^n{x}\,dx} = \frac{\tan^{n - 1}{x}}{n - 1} - \int{\tan^{n - 2}{x}\,dx}}
    Last edited by Prove It; October 2nd 2010 at 09:05 PM.
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  3. #3
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    So \displaystyle{\int{\tan^n{x}\,dx} = \frac{\tan^{n - 1}{x}}{n - 1} - \int{\tan^{n - 1}{x}\,dx}}
    you mean \displaystyle \int\tan^nx dx=\frac{\tan^{n-1} x}{n-1}-\int\tan^{n-2}x dx?
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  4. #4
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    Yes I do
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