u-substitution problem

[mooeypoo]

In one of my solution-guides, they're trying to solve an integral of this form (to later get the root mean square velocity):

 

[math]<v^2>= A \int v^4 \exp{\frac{-Mv^2}{2\tau}}dv[/math]

 

I was thinking of doing this by integration by parts, and trying to reach a gaussian integral or something like that, but I totally miserably failed; it became really complicated. The book solves it by u-substitution, only I don't quite get what they're doing there.

 

They're saying first (and this I understand) that

 

[math]u=\frac{Mv^2}{2\tau}[/math]

 

But then they simply state that

 

[math]v^4 dv=\frac{1}{2}\left( 2\tau/M \right)^{5/2} u^{3/2} du[/math]

 

.... how the hell did they get that? When I translate u to v, I get

 

[math]v^2=\frac{2\tau u}{M}[/math]

 

I tried to switch that into the V and then take a du/dv.. fail. I tried first getting v^4 and then taking derivative:

 

[math]v^4=\frac{4 \tau^2 u^2}{M^2}[/math] and then

 

[math]4v^3dv=\frac{8\tau^2}{M}udu[/math]

 

But this doesn't seem to help me either, since I'm left with v^3dv and not v^4dv..

 

Help? How did they get what they did? After THEY replace it, the u-sub becomes easily solved by Gauss integral (gamma function)..

 

But.. but... how... did they GET this u-substitution to begin with? I am trying to find this forever, and I'm really frustrated.. they seem to think it's.. an "obvious" substitution, which makes this even MORE frustrating.

 

Help will be appreciated!

 

 

=== EDIT ===

 

 

Okay, I feel like an idiot,but I think I got it -- however, since I've never done something like this before, I'm not sure if it's legal?

 

Starting from my last point, I get

 

[math]v^3dv=\frac{2\tau^2}{M}u du[/math]

 

So, now I also know that

 

[math]v=\sqrt{\frac{2\tau u}{M}}[/math]

 

And so I just multiply the above by v...

 

 

[math]v^4dv=\frac{2\tau^2}{M}u\sqrt{\frac{2\tau u}{M}}du[/math]

 

Which seems to be what they're getting... but I'm dealing with derivatives here, (dv/du) -- is it "legal" to multiply by another v pre-derivation? I mean, I differentiate something and THEN multiple by another v, it would change the derivation... so ... is that legal to do here?

 

Thanks.