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Math Help - integral

  1. #1
    Junior Member
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    integral

    <br /> <br />
\int \frac{t^2{dt}}{9+{t^6}}<br /> <br />

    it says to take the the u substitution of  \frac{1}{3}t^3 but how do u get that and is there a simpler way of doing this
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  2. #2
    Super Member
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    the given substitution is a very simple way of doing it.

    u = \frac{1}{3}t^3<br />
    du = t^2 dt

    so the integral becomes

    \int \frac{t^2}{3^2+{3u}^2} \frac{du}{t^2} <br />

    \int \frac{1}{3^2+{3u}^2} du <br />

     = \frac{1}{9} \arctan u<br />

     = \frac{1}{9} \arctan \frac{1}{3}t^3<br />

    I omitted a few step of working. let me know if you can't follow that.

    Bobak
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  3. #3
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    how did you get the very first step of the integral....
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  4. #4
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    as u = \frac{1}{3}t^3 \ \ \Rightarrow \ \ 3u = t^3

    also as du = t^2 dt \ \  \Rightarrow \ \ dt = \frac{du}{t^2}

    so we have

    \int \frac{t^2}{3^2+(t^3)^2} dt we replace dt with \frac{du}{t^2} and (t^3)^2 with (3u)^2

    giving
    \int \frac{t^2}{3^2+{3u}^2} \frac{du}{t^2}<br />

    do you follow the rest ?
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