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Math Help - [SOLVED] Trig Substitution Integral

  1. #1
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    [SOLVED] Trig Substitution Integral

    Evaluate \int \frac{1}{x^2\sqrt{x^2+4}}
    =======================
    Using trig substitution, let x=2\tan{u}, dx=2\sec^2{u}du

    = \int \frac{1}{4\tan^2{u} \cdot \sqrt{4\tan^2{u}+4}} \cdot 2\sec^2{u}du

    = \int \frac{2\sec^2{u}}{\left( 4\tan^2{u} \right) \sqrt{4 ( \tan^2{u}+1) }}du

    = \int \frac{2\sec^2{u}}{\left( 4\tan^2{u} \right) 2 \sqrt{\tan^2{u}+1}}du

    Since \sqrt{\tan^2{u}+1}= \sec{u}, then

    = \int \frac{\sec^2{u}}{4\tan^2{u}\sec{u}} du

    = \frac{1}{4} \int \frac{\sec{u}}{\tan^2{u}} du

    = \frac{1}{4} \int \frac{\csc{u}}{\tan{u}} du

    From here, I do not know how to proceed.

     \csc{u} = \frac{1}{\sin{u}}
     \tan{u} = \frac{\sin{u}}{\sin{u}}

    = \frac{1}{4} \int \frac{\cos{u}}{\sin^2{u}} du

    = \frac{1}{4} \int \frac{\cos{u}}{1-\cos^2{u}} du
    Last edited by Paperwings; December 20th 2009 at 08:34 AM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \frac{\csc u}{\tan u}=\frac{\cos u}{\sin^2u}

    Let \sin u=t
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  3. #3
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    Hello, Paperwings!


    You have: . \frac{1}{4} \int \frac{\csc{u}}{\tan{u}}\,du \;=\;\frac{1}{4}\int \csc u \cot u\,du \;=\;-\frac{1}{4}\csc u + C

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  4. #4
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    Ah, yes. Thank you. So,

    = \frac{1}{4} \int \frac{\cos{u}}{\sin^2{u}} du

    Let  \sin{u} = t ,  \cos{u} = dt

    = \frac{1}{4} \int \frac{1}{t^2}dt

     = - \frac{1}{4t} + C

     = - \frac{1}{4\sin{u}} + C =  - \frac{1}{4}\csc{u} + C

    Since  x = 2\tan{u}

     = - \frac{1}{4} \csc{\arctan{\frac{x}{2}}} + C

    I am a little unsure how to simplify this.

    I picture a triangle with sides x and 2. the hypothenuse is  \sqrt{x^2+4}.

    Ah, ok. it's

     = - \frac{1}{4} \cdot \frac{\sqrt{x^2+4}}{x} + C
    Last edited by Paperwings; December 20th 2009 at 09:14 PM.
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