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Thread: Integral 16

  1. #1
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    Integral 16

    $\displaystyle \int \frac{x^2dx}{\sqrt{(4-x^2)^5}}$
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  2. #2
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    I learnt this from Krizalid aka the Inte-Killer. Try the substitution u = 1/x.

    I wouldn't be surprised if Kriz has another clever substitution.
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  3. #3
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    Trigonometric substitution, $\displaystyle x=2 \sin t$ is a way.

    Another way is below.

    I assume that $\displaystyle x>0$,

    $\displaystyle \frac{x^2}{(4-x^2)^{\frac{5}{2}}} = \frac{x^2}{x^5\left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}} = \frac{1}{x^3 \left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}}$

    $\displaystyle \int \frac{dx}{x^3 \left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}}$

    Let $\displaystyle u = \sqrt{\frac{4}{x^2}-1}$, then $\displaystyle u^2 = \frac{4}{x^2}-1$ and $\displaystyle 2 u~du = -\frac{8}{x^3}~dx$

    Applying the substitution gives us $\displaystyle -\frac{1}{4}\int u^{-4}~du= \frac{1}{12u^3}+C = \frac{1}{12 \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C$

    For $\displaystyle x < 0$, the integral is $\displaystyle -\frac{1}{12 \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C$.

    So we can say that, for any x, the integral is $\displaystyle \frac{\sqrt{x^2}}{12x \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C$
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  4. #4
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    I thought I'd elaborate more on my post.

    $\displaystyle \int \frac{x^2}{\sqrt{(4-x^2)^5}}dx$

    Let $\displaystyle u = \frac{1}{x}$

    $\displaystyle du = -\frac{1}{x^2} dx$

    $\displaystyle - x^2 du = dx$

    Replace:
    $\displaystyle \int \frac{1}{u^4\sqrt{(4-\frac{1}{u^2})^5}}du$

    $\displaystyle \int \frac{1}{u^4\sqrt{(\frac{4u^2-1}{u^2})^5}}du$

    $\displaystyle \int \frac{u^5}{u^4\sqrt{(4u^2-1)^5}}du$

    $\displaystyle \int \frac{u}{\sqrt{(4u^2-1)^5}}du$

    Now it became a matter of simple substition ($\displaystyle z = 4u^2 - 1$).
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  5. #5
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    Hello, Apprentice123!

    Another approach . . .


    $\displaystyle \int \frac{x^2\,dx}{\sqrt{(4-x^2)^5}}\;=\;\int\frac{x^2\,dx}{(4-x)^{\frac{5}{2}}}$

    Let: $\displaystyle x = 2\sin\theta \quad\Rightarrow\quad dx = 2\cos\theta\,d\theta$

    Substitute: .$\displaystyle \int\frac{(2\sin\theta)^2(2\cos\theta\,d\theta)}{( 4\cos^2\!\theta)^{\frac{5}{2}}} \;=\;\int\frac{4\sin^2\!\theta\cdot2\cos\theta\,d\ theta}{32\cos^5\!\theta} \;= \;\frac{1}{4}\int\frac{\sin^2\!\theta}{\cos^4\!\th eta}\,d\theta$

    . . $\displaystyle = \;\frac{1}{4}\int \frac{\sin^2\!\theta}{\cos^2\!\theta}\cdot\frac{1} {\cos^2\!\theta}\,d\theta \;=\;\frac{1}{4}\int\tan^2\!\theta(\sec^2\!\theta\ ,d\theta)$



    Let $\displaystyle u = \tan\theta \quad\Rightarrow\quad du = \sec^2\!\theta\,d\theta$

    Substitute: .$\displaystyle \frac{1}{4}\int u^2\,du \;=\;\frac{1}{12}u^3 + C$


    Back-substitute: .$\displaystyle \frac{1}{12}\tan^3\!\theta + C$


    We had: .$\displaystyle \sin\theta = \frac{x}{2}\quad\Rightarrow\quad \tan\theta = \frac{x}{\sqrt{4-x^2}} $


    Answer: .$\displaystyle \frac{1}{12}\left(\frac{x}{\sqrt{4-x^2}}\right)^3 + C \;=\;\frac{x^3}{12(4-x^2)^{\frac{3}{2}}} + C $

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