# Use Integration by Parts to prove the Reduction Formula

• Feb 19th 2010, 03:03 PM
Tulki
Use Integration by Parts to prove the Reduction Formula
int[(x^2 + a^2)^n] dx =
[x(x^2 + a^2)/(2n + 1)] + [(2na^2)/(2n + 1)]*int[(x^2 + a^2)^(n-1)] dx

Where n =/= -(1/2)

I've done a few other examples of integration by parts just fine, but I can't seem to find selections for u and dv which allow me to solve this.

If anyone can grant me the FIRST step, I'm sure I can take it from there.

Cheers!
• Feb 19th 2010, 05:07 PM
Quote:

Originally Posted by Tulki
int[(x^2 + a^2)^n] dx =
[x(x^2 + a^2)^n/(2n + 1)] + [(2na^2)/(2n + 1)]*int[(x^2 + a^2)^(n-1)] dx

Where n =/= -(1/2)

I've done a few other examples of integration by parts just fine, but I can't seem to find selections for u and dv which allow me to solve this.

If anyone can grant me the FIRST step, I'm sure I can take it from there.

Cheers!

Hi Tulki,

You can let $u=(x^2+a^2)^n,\ dv=dx,\ v=x$

Or let dv=1, v=x.

Integration by parts gives

$\int{\left(x^2+a^2\right)^n}dx=x\left(x^2+a^2\righ t)^n-\int{nx\left(x^2+a^2\right)^{n-1}(2x)}dx$

$=x\left(x^2+a^2\right)^n-2n\int{\left(x^2-a^2+a^2\right)\left(x^2+a^2\right)^{n-1}}dx$

(Adding $a^2-a^2$ to $x^2$ helps solve the integral easily).

$=x\left(x^2+a^2\right)^n-2n\left[\int{\left(x^2+a^2\right)\left(x^2+a^2\right)^{n-1}}dx-\int{a^2\left(x^2+a^2\right)^{n-1}}dx\right]$

$=x\left(x^2+a^2\right)^n-2n\int{\left(x^2+a^2\right)^n}dx+2na^2\int{\left(x ^2+a^2\right)^{n-1}}dx$

$\Rightarrow\ I=x\left(x^2+a^2\right)^n-2nI+2na^2I_{n-1}$

$\Rightarrow\ I(1+2n)=x\left(x^2+a^2\right)^n+2a^2nI_{n-1}$

$\Rightarrow\ I=\frac{x\left(x^2+a^2\right)^n}{1+2n}+\frac{2a^2n }{1+2n}\int{\left(x^2+a^2\right)^{n-1}}dx$
• Feb 19th 2010, 05:11 PM
Tulki
Um... when you quoted the question, you added in a power of n that isn't in the actual question I'm looking at. May have been a mistake, but I'm not sure.
• Feb 19th 2010, 05:13 PM