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Math Help - Integration Challenge

  1. #1
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    Cool Integration Challenge

    Prove the integral of 1/(x^3+8) between the x terminals 4 and 1 equals (pie root 3)/36
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  2. #2
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    Quote Originally Posted by nath_quam
    Prove the integral of 1/(x^3+8) between the x terminals 4 and 1 equals (pie root 3)/36
    You can express,
    \frac{1}{x^3+8}
    As,
    \frac{1}{(x+2)(x^2-2x+4)}
    Then decompose this fraction as,
    \frac{A}{x+2}+\frac{Bx+C}{x^2-2x+4}

    When fully decomposed gives,
    <br />
\frac{2}{3}\cdot \int^4_1 \frac{1/8}{x+2}+\frac{-1/8x+1/2}{x^2-2x+4} dx

    The first summand calculates to,
    \left \int^4_1 \frac{1/8}{x+2}dx=\ln |x+2| \right|^4_1=\ln 6-\ln 3=\ln 2
    Multiply by constant 1/8 and 2/3,
    (1/12)\ln 2.

    The second summand need manipulation,
    \int^4_1 \frac{-1/8x+1/2}{x^2-2x+4} dx=-\frac{1}{8}\int^4_1 \frac{(x-1)-3}{(x-1)^2+3} dx

    Apply a linear substitution, u=x-1
     \int^3_0 \frac{u-3}{u^2+3}du
    Express as,
    \int^3_0 \frac{u}{u^2+3}-\frac{1}{(\frac{u}{\sqrt{3}})^2+1} du

    Yields, (first summand is natural logarithm in disuise, the second is the arctanget),
    \left \frac{1}{2}\ln|u^2+3| \right|^3_0 - \sqrt{3} \left \tan^{-1} \left( \frac{u}{\sqrt{3}} \right) \right|^3_0
    The first summand gives,
    \frac{1}{2}\ln 12-\frac{1}{2}\ln 3=\frac{1}{2} (\ln 12-\ln 3)=\frac{1}{2}\ln 4=\ln 4^{1/2}=\ln 2
    The second summand gives,
    \sqrt{3} \tan^{-1} \left( \frac{3}{\sqrt{3}} \right) -\sqrt{3} \tan^{-1} (0)=\frac{\pi\sqrt{3}}{3}
    The next step is to subtract these summands,
    \ln 2-\frac{\pi\sqrt{3}}{3}
    Next is to mutiply this by -1/8 thus,
    -(1/8)\ln 2+\frac{\pi\sqrt{3}}{24}
    Now mutiply to 2/3,
    -(1/12)\ln 2+\frac{\pi\sqrt{3}}{36}
    And add your very first summand from the beginning,
    thus,
    (1/12)\ln 2-(1/12)\ln 2+\frac{\pi\sqrt{3}}{36}=\frac{\pi\sqrt{3}}{36}

    \mathbb{Q}.\mathbb{E}.\mathbb{D}
    Last edited by ThePerfectHacker; June 6th 2006 at 01:49 PM.
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