1. ## 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.

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

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

2. Originally Posted by bendevlin
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.

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

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, $\displaystyle r = a \tan \theta$

do you know how to do trig substitution? if not, we can look for another way

3. Originally Posted by bendevlin
$\displaystyle \int \frac {r^2 dr}{(a^2+r^2)^{3/2}}$,
Step #1: The make-up.

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

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

Step #2: Integrate by separately.

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

$\displaystyle \int {\frac{1} {{\sqrt {r^2 + a^2 } }}\,dr} = \ln \left| {r + \sqrt {r^2 + a^2 } } \right| + k_1 .$

For the second one, substitute $\displaystyle r=\frac1v,$

$\displaystyle \int {\frac{1} {{\left( {r^2 + a^2 } \right)\sqrt {r^2 + a^2 } }}\,dr} = - \int {\frac{v} {{\left( {1 + a^2 v^2 } \right)\sqrt {1 + a^2 v^2 } }}\,dv}.$

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

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

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

So, we happily get

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

Mission Accomplished