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Math Help - Substitution

  1. #1
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    Substitution

    Good morning,

    I am looking at the following function, trying to understand what value I should be substituting u for.

    \int\frac{{\ln}(cos^{-1}x)dx}{cos^{-1}x\sqrt{1-x^2}}

    I am coming up with:

    u=\ln(cos^{-1}x)

    du=\frac{1}{cos^{-1}x}

    I believe this should give me:

    \int{ududdu}



    More importantly, I am really struggling with the "why" part of this substitution. I must be missing something obvious.
    Last edited by MechEng; May 26th 2010 at 07:43 AM. Reason: LaTex errors... among other things
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  2. #2
    MHF Contributor ebaines's Avatar
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    Not quite right. If you set

    <br />
u=ln(cos^{-1}x)<br />

    Then take the derivative - don't forget to apply the chain rule:
    <br />
\frac {du} {dx} = \frac 1 {cos^{-1} x} \cdot \frac {d(cos^{-1} x)} {dx} = \frac 1 {cos^{-1} x} \cdot \frac {-1} {\sqrt {1-x^2}}<br />
    <br />
du = \frac {-dx} {cos^{-1} x\sqrt {1-x^2}}<br />

    So now you have:

    <br />
\int \frac {ln(cos^{-1}x ) dx} {cos^{-1}x \sqrt {1 - x^2}} = -\int u du =-\frac 1 2 u^2 + C = -\frac 1 2 (ln(cos^{-1}x)^2 + C<br />

    You could have also done this by setting u = cos^{-1} x:

    <br />
u=cos^{-1}x<br />
    <br />
\frac {du} {dx} = \frac {d(cos^{-1} x)} {dx} = \frac {-1} {\sqrt {1-x^2}}<br />
    <br />
du = \frac {-1} {\sqrt {1-x^2}}dx<br />
    <br />
\int \frac {ln(cos^{-1}x ) dx} {cos^{-1}x \sqrt {1 - x^2}} = - \int \frac {ln(u)} u du = - \frac 1 2 (ln(u))^2 + C = - \frac 1 2 (ln(cos^{-1}x))^2 + C<br />

    Always happy to help out a fellow ME!
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