Reduction Method Question

**shonenhikada** · April 22nd 2009, 02:03 AM

http://i41.tinypic.com/ml7ez6.jpg

Help please.

**Danny** · April 22nd 2009, 05:07 AM

First off, there's a typo (an n is missing)

$\int(x^2+a^2)^ndx$ if $u = (x^2+a^2)^n,\;\; dv= dx$ so $du = 2 n x (x^2+a^2)^{n-1},\;\;v = x$ so

$I_n = \int(x^2+a^2)^ndx$
$= x(x^2+a^2)^n - 2n \int x^2 (x^2+a^2)^{n-1}dx$
$= x(x^2+a^2)^n - 2n \int (x^2+a^2-a^2) (x^2+a^2)^{n-1}dx$
$= x(x^2+a^2)^n - 2n \int (x^2+a^2) (x^2+a^2)^{n-1}dx + 2na^2 \int (x^2+a^2)^{n-1}dx$
$I_n = x(x^2+a^2)^n - 2n \underbrace{\int (x^2+a^2)^{n}dx}_{I_n} + 2na^2 \underbrace{\int (x^2+a^2)^{n-1}dx}_{I_{n-1}}$
so

$(1+2n) I_n = x(x^2+a^2)^n +2na^2 I_{n-1}$

from we obtain your result.

**shonenhikada** · April 22nd 2009, 05:16 AM

Thank you for your help but how did you know their was a typo. I though I had to manipulate the x^2 + a^2 to get it into a high power like saying x^2 + a^2 = (x^2 + a^2) ^n * (x^2 + a^2)^1-n

**Danny** · April 22nd 2009, 05:35 AM

Originally Posted by shonenhikada

Thank you for your help but how did you know their was a typo. I though I had to manipulate the x^2 + a^2 to get it into a high power like saying x^2 + a^2 = (x^2 + a^2) ^n * (x^2 + a^2)^1-n

You circled a term and called it $I_{n-1}$ (I change it for this conversation). This suggests that there is an $I_{n}$ . Second, put $a=0,\;\;n=2$ - your formula doesn't work!

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