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Math Help - Integral problem

  1. #1
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    Integral problem

    I am trying to solve the integral:

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


    This is what I did:

    u = tan(x)
    x = arctan(u)
    dx = (1+u^2) du

    So that the integral is:

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

    Then I did polynomial division, and got that:


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

    \int\frac{1}{2}u^2 + \frac{1}{4} -\frac{1}{4}\frac{1}{2u^2+1} du

    But the according to Wolfram, the answer that I should have got from the integral I started solving, is not the same as the answer to this new integral.

    What mistake did I do?
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  2. #2
    MHF Contributor harish21's Avatar
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    you are doing it right..after integrating, you have to substitute back the value of u, which is tanx.

    Edit: A correction is needed in the step where you found dx. Look at the post below.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Is...

    \displaystyle u= \tan x \implies x = \tan^{-1} u \implies \frac{dx}{du}= \frac{1}{1+u^{2}}

    Kind regards

    \chi \sigma
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  4. #4
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    You might go an other way, or whatever we like to cal it. If we massage the trigonometric functions out of tan²x instead.

    \displaystyle tan²x = \frac{sin²x}{cos²x} = \frac{\frac{1-cos 2x}{2}}{\frac{1+cos2x}{2}} = \frac{1-cos 2x}{1+cos2x}

    In your integral

    \displaystyle \frac{tan²x}{1+2tan²x} = \frac{\frac{1-cos 2x}{1+cos2x}}{1+2 \left( \frac{1-cos 2x}{1+cos2x} \right)} = \frac{\frac{1-cos 2x}{1+cos2x}}{\frac{1+cos2x + 2-2cos 2x}{1+cos2x} } = \frac{1-cos 2x}{3-cos2x} } = \frac{2-2 + 1-cos 2x}{3-cos2x}  = 1 - \frac{2}{3-cos2x}

    Try solving \int 1 - \frac{2}{3-cos2x} dx and see if it computes.
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  5. #5
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    To make things easier:

    \displaystyle \int\frac{\tan^2{x}}{1+2\tan^2{x}}\;{dx} = \frac{1}{2}\int\frac{2\tan^2{x}-1+1}{1+2\tan^2{x}}\;{dx} = \frac{1}{2}\;{x}-\frac{1}{2}\int\frac{1}{1+2\tan^2{x}}\;{dx}.

    Still go on with [LaTeX ERROR: Convert failed] .
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