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Math Help - Integration

  1. #1
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    Integration

    Hey guys,

    I've got this problem in my exam that seems to come up a lot, I'm really stuck on where to go next.

    \int \frac {r^2 dr}{(a^2+r^2)^{3/2}},

    I thought about substitution and integration by parts but can't figure out a way to make the r's cancel out.

    If anyone has some advice on how to start this it would be greatly apreciated!

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bendevlin View Post
    Hey guys,

    I've got this problem in my exam that seems to come up a lot, I'm really stuck on where to go next.

    \int \frac {r^2 dr}{(a^2+r^2)^{3/2}},

    I thought about substitution and integration by parts but can't figure out a way to make the r's cancel out.

    If anyone has some advice on how to start this it would be greatly apreciated!

    Thanks
    you could use a trig substitution, r = a \tan \theta

    do you know how to do trig substitution? if not, we can look for another way
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by bendevlin View Post
    \int \frac {r^2 dr}{(a^2+r^2)^{3/2}},
    Step #1: The make-up.

    \int {\frac{{r^2 }}<br />
{{\left( {r^2  + a^2 } \right)^{3/2} }}\,dr}  = \int {\frac{{r^2  + a^2  - a^2 }}<br />
{{\left( {r^2  + a^2 } \right)^{3/2} }}\,dr}.

    So \int {\frac{{r^2 }}<br />
{{\left( {r^2  + a^2 } \right)^{3/2} }}\,dr}  = \int {\frac{1}<br />
{{\sqrt {r^2  + a^2 } }}\,dr}  - a^2 \int {\frac{1}<br />
{{\left( {r^2  + a^2 } \right)\sqrt {r^2  + a^2 } }}\,dr} .


    Step #2: Integrate by separately.

    For the first integral, set u = r + \sqrt {r^2  + a^2 },

    \int {\frac{1}<br />
{{\sqrt {r^2  + a^2 } }}\,dr}  = \ln \left| {r + \sqrt {r^2  + a^2 } } \right| + k_1 .

    For the second one, substitute r=\frac1v,

    \int {\frac{1}<br />
{{\left( {r^2  + a^2 } \right)\sqrt {r^2  + a^2 } }}\,dr}  =  - \int {\frac{v}<br />
{{\left( {1 + a^2 v^2 } \right)\sqrt {1 + a^2 v^2 } }}\,dv}.

    One more substitution according to \varphi ^2  = 1 + a^2 v^2,

    \int {\frac{1}<br />
{{\left( {r^2  + a^2 } \right)\sqrt {r^2  + a^2 } }}\,dr}  =  - \frac{1}<br />
{{a^2 }}\int {\frac{1}<br />
{{\varphi ^2 }}\,d\varphi }  = \frac{1}<br />
{{a^2 \varphi }} + k_2 .


    Step #3: Back substitute, combine constants and we're done.

    So, we happily get

    \int {\frac{{r^2 }}<br />
{{\left( {r^2  + a^2 } \right)^{3/2} }}\,dr}  = \ln \left| {r + \sqrt {r^2  + a^2 } } \right| - \frac{r}<br />
{{\sqrt {r^2  + a^2 } }} + k.


    Mission Accomplished
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