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Thread: Integral of 1/(1+4x^2)

  1. #1
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    Integral of 1/(1+4x^2)

    Problem:

    $\displaystyle \int \frac{dx}{1+4x^2}dx$

    Attempt:

    $\displaystyle 4x = tan(\theta)$
    $\displaystyle dx = \frac{d\theta}{cos^2(\theta)}$

    $\displaystyle \int \frac {d\theta}{(1+tan^2(\theta))*cos^2(\theta)}$

    $\displaystyle \int \frac {d\theta}{(1 + \frac{sin^2(\theta)}{cos^2(\theta)})*cos^2(\theta) }$

    $\displaystyle \int \frac{d\theta}{cos^2(\theta)+sin^2(\theta)}$

    $\displaystyle \int d\theta$

    $\displaystyle \theta + C$

    My Answer:
    $\displaystyle arctan(4x) + C$

    Correct Answer:
    $\displaystyle
    \frac{1}{2} arctan(2x) + C
    $
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  2. #2
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    correction ...

    $\displaystyle 1 + 4x^2 = 1 + (2x)^2
    $

    $\displaystyle 2x = \tan{\theta}$
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  3. #3
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    Quote Originally Posted by Fallen186 View Post
    Problem:

    $\displaystyle \int \frac{dx}{1+4x^2}dx$

    Attempt:

    $\displaystyle 4x = tan(\theta)$
    $\displaystyle dx = \frac{d\theta}{cos^2(\theta)}$

    $\displaystyle \int \frac {d\theta}{(1+tan^2(\theta))*cos^2(\theta)}$

    $\displaystyle \int \frac {d\theta}{(1 + \frac{sin^2(\theta)}{cos^2(\theta)})*cos^2(\theta) }$

    $\displaystyle \int \frac{d\theta}{cos^2(\theta)+sin^2(\theta)}$

    $\displaystyle \int d\theta$

    $\displaystyle \theta + C$

    My Answer:
    $\displaystyle arctan(4x) + C$

    Correct Answer:
    $\displaystyle
    \frac{1}{2} arctan(2x) + C
    $
    Close. Try $\displaystyle 2x = \tan \theta$
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  4. #4
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    If $\displaystyle 4x=\tan\,\theta$, then $\displaystyle x=\frac{1}{4}\tan\,\theta$ and $\displaystyle dx=\frac{1}{4\,\cos^2\theta}\,d\theta$.
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  5. #5
    Senior Member Twig's Avatar
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    $\displaystyle \int \frac{1}{1+4x^2}\, dx = \int \frac{1}{1+(2x)^{2}} \, dx = \frac{1}{2}\cdot arctan(2x)+C $
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