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Math Help - Integral of 1/(1+4x^2)

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

    Problem:

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

    Attempt:

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

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

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

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

    \int d\theta

    \theta + C

    My Answer:
    arctan(4x) + C

    Correct Answer:
     <br />
\frac{1}{2} arctan(2x) + C<br />
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  2. #2
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    correction ...

    1 + 4x^2 = 1 + (2x)^2<br />

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

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

    Attempt:

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

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

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

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

    \int d\theta

    \theta + C

    My Answer:
    arctan(4x) + C

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