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Math Help - antiderivative w/ tan

  1. #1
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    antiderivative w/ tan

    Having trouble with this

    find the antiderivative of
    \frac{1}{81z^2+16}
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  2. #2
    Super Member Quacky's Avatar
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    Re: antiderivative w/ tan

    We have something of the form \frac{1}{a^2+x^2} which is a standard form that you can directly integrate.

    If you don't want to do that, then we're going to have to do an icky substitution.

    Let z=\frac{4}{9}tan(x)

    Then dz=\frac{4}{9}sec^2{x}~dx

    I=\int\frac{1}{81z^2+16}dz

    =\int\frac{\frac{4}{9}sec^2{x}}{81(\frac{16}{81}ta  n^2(x))+16}dx

    I think it should work out. Good luck!
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