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Thread: Simple Integral

  1. #1
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    Simple Integral

    So i'm having trouble with a integral that i've been seeing alot lately..It goes like $\displaystyle \frac{1}{A+x^2}$ where A could be anything that is not 1...Now the solution ends up being arctan. However i've been struggling to figure out how to substitute in U to get the right answer...for example $\displaystyle \frac{1}{8+x^2}$ . What should i substitute?
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  2. #2
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    In the integral

    $\displaystyle \int\frac{dx}{{a^2+x^2}}$

    By Trigonometric Substitution:

    $\displaystyle x=a\tan(\theta),\quad dx=a\sec^2(\theta)\,d\theta$

    $\displaystyle \theta=\arctan\left(\frac{x}{a}\right)$
    so that the integral becomes

    $\displaystyle
    \qquad \int\frac{dx}{{a^2+x^2}}$
    = [Math]\int\frac{a\sec^2(\theta)\,d\theta}{{a^2+a^2\tan^2 (\theta)}}[/tex]
    = $\displaystyle \int\frac{a\sec^2(\theta)\,d\theta}{{a^2(1+\tan^2( \theta))}}$
    = $\displaystyle \int \frac{a\sec^2(\theta)\,d\theta}{{a^2\sec^2(\theta) }}$
    $\displaystyle = \int \frac{d\theta}{a} = \frac{\theta}{a}+C $
    $\displaystyle = \frac{1}{a} \arctan \left(\frac{x}{a}\right)+C$


    (provided ''a'' > 0).

    The long and short of it is when you have an integral of the form $\displaystyle \int\frac{dx}{{a^2+x^2}}$ you get an result of the form: $\displaystyle \frac{1}{a} \arctan \left(\frac{x}{a}\right)+C$

    It's a good one to memorize.
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