Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By votan

Math Help - Help with integral

  1. #1
    Junior Member
    Joined
    Nov 2012
    From
    Mexico
    Posts
    53
    Thanks
    7

    Help with integral

    Hi, I'm having some trouble with the following integral:

    i \int sin(t) e^{it} dt

    Making u=sin(t), then du=cos(t)
    And if dv=e^{it}, then v=-ie^{it}

    So

    i \int sin(t) e^{it} dt= i(-ie^{it}sin(t) - \int -ie^{it}cos(t) dt)= =e^{it}sin(t) -\int e^{it}cos(t) dt

    Now, for the integral of e^{it}cos(t) dt:

    u=cos(t), du=-sin(t)
    dv=e^{it}, v=-ie^{it}

    So
    \int e^{it}cos(t)dt= -ie^{it}cos(t)- \int -ie^{it}(-sin(t))dt= -ie^{it}cos(t)- i\int e^{it}sin(t)dt

    So
    i \int sin(t) e^{it} dt= e^{it}sin(t) +ie^{it}cos(t)+ i\int e^{it}sin(t)dt

    So I guess integration by parts won't do it here. How can I solve it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Sep 2013
    From
    USA
    Posts
    255
    Thanks
    114

    Re: Help with integral

    convert i*e(it) to i*(cos(t) + i*sin(t)

    If you want to use by parts, set i*e(it)dt = du and sin(t)
    Last edited by votan; October 10th 2013 at 02:01 PM.
    Thanks from Cesc1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2012
    From
    Mexico
    Posts
    53
    Thanks
    7

    Re: Help with integral

    Quote Originally Posted by votan View Post
    convert i*e(it) to i*(cos(t) + i*sin(t)

    If you want to use by parts, set i*e(it)dt = du and sin(t)
    Thanks, I'll try that. Is there other way besides by parts?

    Edit: Just figured it's better to express sin t as (1/2)(-ieit+ie-it) and the integral becomes so much easier.
    Last edited by Cesc1; October 10th 2013 at 02:53 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 07:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum