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Math Help - Integrating (Inverse Trigonometric Functions)

  1. #1
    Junior Member xfriendsonfirex's Avatar
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    Integrating (Inverse Trigonometric Functions)

    Maybe I'm missing something obvious here, but I can't seem to integrate this. I'm just about positive I need to use arctan, but I can't get a^2+u^2 in the denominator.

    \int \frac{1}{3 + (x-2)^2}dx

    Any help is appreciated!
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  2. #2
    Eater of Worlds
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    Let u=x-2, \;\ du=dx

    \int\frac{1}{3+u^{2}}du

    \frac{1}{3}\int\frac{1}{1+(\frac{u}{\sqrt{3}})^{2}  }du

    Now, use arctan.
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  3. #3
    Junior Member xfriendsonfirex's Avatar
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    Do you mind looking through to see if I did this correctly?

    Using the formula for arctan:

    \int \frac{du}{a^2+u^2}=\frac{1}{a}arctan\frac{u}{a}+C

    And letting a=1, u=\frac{x-2}{\sqrt{3}}, and du=\frac{\sqrt{3}}{3}:

    \frac{1}{3}arctan\frac{x-2}{\sqrt{3}}+C

    Do you see anything I missed?
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  4. #4
    Eater of Worlds
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    Looks good, except I think it should be \frac{1}{\sqrt{3}}tan^{-1}(\frac{x-2}{\sqrt{3}})+C
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