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Thread: [SOLVED] Trig Substitution Integral

  1. #1
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    [SOLVED] Trig Substitution Integral

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

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

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

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

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

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

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

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

    From here, I do not know how to proceed.

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

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

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

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


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

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

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

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

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

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

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

    Since $\displaystyle x = 2\tan{u} $

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

    I am a little unsure how to simplify this.

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

    Ah, ok. it's

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