Results 1 to 5 of 5

Math Help - Integration by reduction

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    2

    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?

    Can someone please help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NewOrcOrder View Post
    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?

    Can someone please help?
    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: \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:

    \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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2007
    Posts
    2
    Quote Originally Posted by Jhevon View Post
    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: \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:

    \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.

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


    and I am stuck....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NewOrcOrder View Post
    Thank you, but I'm currently learning averaged equations and I need to break down my equation as in the example below.

    <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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration Reduction; Formulae
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 23rd 2011, 08:37 PM
  2. Reduction method of integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 2nd 2010, 10:05 PM
  3. Integration reduction formulae
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 3rd 2010, 09:35 PM
  4. Integration - Reduction Method: PLEASE HELP!!!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 9th 2010, 08:38 PM
  5. Integration using Reduction formulae
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 6th 2007, 11:05 AM

Search Tags


/mathhelpforum @mathhelpforum