Integration

**bendevlin** · January 15th 2008, 05:02 PM

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

**Jhevon** · January 15th 2008, 05:06 PM

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.

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

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

**Krizalid** · January 15th 2008, 05:22 PM

Originally Posted by bendevlin

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

Step #1: The make-up.

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

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

Step #2: Integrate by separately.

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

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

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

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

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

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

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

So, we happily get

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

Mission Accomplished

Thread: Integration

LinkBack

Thread Tools

Display

Integration

Similar Math Help Forum Discussions

Integration help please these are three small even basic integration problems

sketch the region of integration and change the order of integration.

Definite Integration - Double integration by parts

Simple Integration problem: Trigonometric Integration

Tricky integration, possibly Integration Factor

Search Tags