Integration by reduction

• Sep 25th 2007, 05:39 PM
NewOrcOrder
Integration by reduction
Hi

I need to integrate sin^3 cos and want to use the reduction formula to do it.

since sin^2(x) = 1-cos^2(x)

it becomes sinx (cosx - cos^3 x).. but I don't know how to continue on.

Is there anyway to get rid of the multiplication?

• Sep 25th 2007, 05:53 PM
Jhevon
Quote:

Originally Posted by NewOrcOrder
Hi

I need to integrate sin^3 cos and want to use the reduction formula to do it.

since sin^2(x) = 1-cos^2(x)

it becomes sinx (cosx - cos^3 x).. but I don't know how to continue on.

Is there anyway to get rid of the multiplication?

i don't see your problem? just plug in the required values into the reduction formula (this is by the way too much work, i'd use a simple substitution to do this).

recall: $\displaystyle \int \sin^n u \cos^m u ~du = - \frac {\sin^{n - 1} u \cos^{m + 1} u}{n + m} + \frac {n - 1}{n + m} \int \sin^{n - 2} u \cos^m u~du$

or alternately:

$\displaystyle \int \sin^n u \cos^m u ~du = \frac {\sin^{n + 1} u \cos^{m - 1} u}{n + m} + \frac {m - 1}{n + m} \int \sin^n u \cos^{m - 2} u~du$

as i said, this is way more trouble than it's worth. if you are not required to use reduction, do the integral with substitution
• Sep 25th 2007, 06:05 PM
NewOrcOrder
Quote:

Originally Posted by Jhevon
i don't see your problem? just plug in the required values into the reduction formula (this is by the way too much work, i'd use a simple substitution to do this).

recall: $\displaystyle \int \sin^n u \cos^m u ~du = - \frac {\sin^{n - 1} u \cos^{m + 1} u}{n + m} + \frac {n - 1}{n + m} \int \sin^{n - 2} u \cos^m u~du$

or alternately:

$\displaystyle \int \sin^n u \cos^m u ~du = \frac {\sin^{n + 1} u \cos^{m - 1} u}{n + m} + \frac {m - 1}{n + m} \int \sin^n u \cos^{m - 2} u~du$

as i said, this is way more trouble than it's worth. if you are not required to use reduction, do the integral with substitution

Thank you, but I'm currently learning averaged equations and I need to break down my equation as in the example below.

$\displaystyle <cos^2sin^4> = <(1-sin^2)sin^4> = <sin^4> -<sin^6> = 3/8 - 15/48$

and I am stuck....
• Sep 25th 2007, 06:09 PM
Jhevon
Quote:

Originally Posted by NewOrcOrder
Thank you, but I'm currently learning averaged equations and I need to break down my equation as in the example below.

$\displaystyle <cos^2sin^4> = <(1-sin^2)sin^4> = <sin^4> -<sin^6> = 3/8 - 15/48$

and I am stuck....

i'm sorry, but i have no idea what you are talking about. and you can't just write sin and cos, you have to have the sine of something and the cosine of something. so write sin(x) and cos(x) or whatever
• Sep 25th 2007, 06:53 PM
Krizalid
Do we have to kill a fly with a Grand Cannon?

I think it's extremely unnecessary to use a big reduction formula for such integral.

Do what Jhevon says, make a little substitution.

--

You should apply the reduction formulas when the integral be really messy.