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Math Help - Integration of f(x)^n Where n is -ve fraction

  1. #1
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    Integration of f(x)^n Where n is -ve fraction

    I am trying to integrate the following problem with limit +D and -D

    \int{\frac{1}{(z^2 + r^2)^\frac{3}{2}}\,dz}

    Now intuition tells me to treat it like

    \int{(z^2 + r^2)^\frac{-3}{2}\,dz}

    Here is where I'm going wrong - the way I'm integrating gives me

    \frac{(2z)(z^2 + r^2)^\frac{-1}{2}}{\frac{-1}{2}}

    \frac{-4z}{(z^2 + r^2)^\frac{-1}{2}}

    But if I differentiate that I get

    \frac{4r^2}{(z^2 + r^2)^\frac{3}{2}}

    So have randomly created a factor of 4r^2. Therefore I'm doing something wrong but I can't seem to get a rule to fit the situation.

    Differentiating

    \int{\frac{z}{r^2(z^2 + r^2)^\frac{1}{2}}\,dz}

    Returns me to the original problem

    \frac{1}{(z^2 + r^2)^\frac{3}{2}}

    But that would imply that the basic integral is divided by r^2 despite the fact that for all intents and purposes r^2 is a constant.

    Can anyone explain, as I think I've horribly confused myself about something not that complex?
    Last edited by M1ke; January 11th 2009 at 07:40 AM.
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  2. #2
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    Quote Originally Posted by M1ke View Post
    I am trying to integrate the following problem with limit +D and -D

    \frac{1}{(z^2 + r^2)^\frac{3}{2}}

    Now intuition tells me to treat it like

    (z^2 + r^2)^\frac{-3}{2}

    Here is where I'm going wrong - the way I'm integrating gives me

    \frac{(2z)(z^2 + r^2)^\frac{-1}{2}}{\frac{-1}{2}}

    \frac{-4z}{(z^2 + r^2)^\frac{-1}{2}}

    But if I differentiate that I get

    \frac{4r^2}{(z^2 + r^2)^\frac{3}{2}}

    So have randomly created a factor of 4r^2. Therefore I'm doing something wrong but I can't seem to get a rule to fit the situation.

    Differentiating

    \frac{z}{r^2(z^2 + r^2)^\frac{1}{2}}

    Returns me to the original problem

    \frac{1}{(z^2 + r^2)^\frac{3}{2}}

    But that would imply that the basic integral is divided by r^2 despite the fact that for all intents and purposes r^2 is a constant.

    Can anyone explain, as I think I've horribly confused myself about something not that complex?
    Which variable are you integrating with respect to?

    Is it \int{\frac{1}{(z^2 + r^2)^\frac{3}{2}}\,dz} or \int{\frac{1}{(z^2 + r^2)^\frac{3}{2}}\,dr}?
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  3. #3
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    sorry - all by dz.

    Will amend the original post.
    Last edited by ThePerfectHacker; January 11th 2009 at 11:07 AM.
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  4. #4
    Senior Member DeMath's Avatar
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    Try z = r \cdot \sinh \left( t \right), dz = r \cdot \cosh \left( t \right)dt.
    Last edited by DeMath; January 11th 2009 at 08:10 AM.
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  5. #5
    Senior Member DeMath's Avatar
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    Try to use the hyperbolic functions.

    Then

    \int {\frac{{dz}}{{{{\left( {{z^2} + {r^2}} \right)}^{{3 \mathord{\left/{\vphantom {3 2}} \right.\kern-\nulldelimiterspace} 2}}}}}}  = \left\{ \begin{gathered}z = r \cdot \sinh \left( t \right), \hfill \\dz = r \cdot \cosh \left( t \right)dt \hfill \\ \end{gathered}  \right\} = \frac{1}{{{r^2}}}\int {\frac{{dt}}{{{{\cosh }^2}\left( t \right)}}}  = \frac{1}{{{r^2}}}\tanh \left( t \right) + C =

    = \frac{1}{{{r^2}}}\frac{{\sinh \left( t \right)}}<br />
{{\cosh \left( t \right)}} + C = \frac{1}<br />
{{{r^2}}}\frac{{\sinh \left( {{\text{arcsinh}}\left( {{z \mathord{\left/<br />
 {\vphantom {z r}} \right.\kern-\nulldelimiterspace} r}} \right)} \right)}}<br />
{{\cosh \left( {{\text{arcsinh}}\left( {{z \mathord{\left/<br />
 {\vphantom {z r}} \right.\kern-\nulldelimiterspace} r}} \right)} \right)}} + C =  \frac{1}{{{r^2}}}\frac{{{z \mathord{\left/<br />
{\vphantom {z r}} \right.\kern-\nulldelimiterspace} r}}}<br />
{{\sqrt {1 + {{{z^2}} \mathord{\left/{\vphantom {{{z^2}} {{r^2}}}} \right.\kern-\nulldelimiterspace} {{r^2}}}} }} + C = \frac{z}<br />
{{{r^2}\sqrt {{z^2} + {r^2}} }} + C.
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  6. #6
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    People tend to be more familar with the trig functions so I might recommend

    z = r \tan \theta

    BTW - it reduces to a nice integral afterwards.
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