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Thread: 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 $\displaystyle a^2+u^2$ in the denominator.

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

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

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

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

    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:

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

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

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

    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 $\displaystyle \frac{1}{\sqrt{3}}tan^{-1}(\frac{x-2}{\sqrt{3}})+C$
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