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Math Help - Integration problem involving trig

  1. #1
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    Integration problem involving trig

    The question:

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

    My attempt:
    Let u = tanx
    du = sec^2x dx

    \int{u^{n-2} \ du}

    = (n-2)u^{n-1}
    = (n-2)tan^{n-1}x

    The answer in my text is:

    \frac{tan^{n-1}x}{n-1}

    Where have I gone wrong? Thanks.
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  2. #2
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    \int u^{n-2}du = \frac{u^{n-1}}{n-1}+C
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  3. #3
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    Oops! >_< Thanks.
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