Reduction formula/Integration by parts problem

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Homework Statement


The problem asks me to: Determine a reduction formula for the stated integral (See Picture and problem at:

http://garciarussellchem.angelfire.com/Photo/

Integration_by_parts.jpg

Please help me out with this problem. I don't understand why they do not include an extra r in the denominator when integrating dv. I also don't understand what purpose the r^(n-1) serves.

This is a bit confusing.

Thank you.
 
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They are using u = r^{n-1} and dv = re^{-ar^2}dr then doing integration by parts with those...
 
This means that v must equal"

v= (re^-(ar^2))/-2ar in order for us to get the dv back right?

THANK YOU FOR THE QUICK REPLY :) !
 
No prob.

If you differentiate \frac{-1}{2a}e^{-ar^2} you get re^{-ar^2}, so that's why v = \frac{-1}{2a}e^{-ar^2}
 
Thanks a bunch.
This site rocks.

(thank you)^(10)
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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