Results 1 to 5 of 5

Math Help - how do i find this integral?

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    71

    how do i find this integral?

    integral(sinte^(6/489)(t))
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2007
    Posts
    71
    I know that I cannot use integration by parts since it always loops.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by lord12 View Post
    I know that I cannot use integration by parts since it always loops.
    Is this your integral: \int e^{\frac{6}{489}t}\sin t\,dt??

    If so you can use integration by parts. The key to this one is that you apply integration by parts twice. After you apply it twice, you'll see the original integral reappear.

    I.e. you'll see after applying IBP twice that you will end up with \int e^{\frac{6}{489}t}\sin t\,dt= \text{ stuff } + \text{ constants }\cdot \int e^{\frac{6}{489}t}\sin t\,dt

    Then you treat \int e^{\frac{6}{489}t}\sin t\,dt as a variable and then solve for it.

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, lord12!

    Yes, it "loops" when doing it by-parts, but that's what we want!


    \int e^{\frac{6}{489}t}\sin t\,dt . Why did they use such an ugly fraction?
    Let k = \tfrac{6}{489}

    We have: . I \;=\;\int e^{kt}\sin t\,dt


    . . By parts: . \begin{array}{ccccccc}u &=& \sin t && dv &=& e^{kt}dt \\ du &=& \cos t\,dt & & v &=&\frac{1}{k}e^{kt} \end{array}

    And we have: . I \;=\;\frac{1}{k}\,e^{kt}\sin t - \frac{1}{k}\int e^{kt}\cos t\,dt


    . . By parts: . \begin{array}{ccccccc}u &=& \cos t && dv &=&e^{kt}dt \\ du&=& -\sin t\,dt && v &=&\frac{1}{k}e^{kt} \end{array}

    And we have: . I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k}\left[\frac{1}{k}e^{kt}\cos t + \frac{1}{k}\int e^{kt}\sin t\,dt\right]

    . . . . . . . . . . I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k^2}e^{kt}\cos t - \frac{1}{k^2}\underbrace{\int e^{kt}\sin t\,dt}_{\text{This is }I} + C


    We have: . I + \frac{1}{k^2}I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k^2}e^{kt}\cos t + C

    Factor: . \frac{k^2+1}{k^2}\,I \;=\;\frac{e^{kt}}{k^2}(k\sin t -\cos t) + C

    Then: . I \;=\;\frac{k^2}{k^2+1}\cdot\frac{e^{kt}}{k^2}(k\si  n t - \cos t) + C

    Therefore: . I \;=\;\frac{e^{kt}}{k^2+1}(k\sin t - \cos t) + C

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2008
    Posts
    154
    Quote Originally Posted by Soroban View Post
    Hello, lord12!

    Yes, it "loops" when doing it by-parts, but that's what we want!


    Let k = \tfrac{6}{489}

    We have: . I \;=\;\int e^{kt}\sin t\,dt


    . . By parts: . \begin{array}{ccccccc}u &=& \sin t && dv &=& e^{kt}dt \\ du &=& \cos t\,dt & & v &=&\frac{1}{k}e^{kt} \end{array}

    And we have: . I \;=\;\frac{1}{k}\,e^{kt}\sin t - \frac{1}{k}\int e^{kt}\cos t\,dt


    . . By parts: . \begin{array}{ccccccc}u &=& \cos t && dv &=&e^{kt}dt \\ du&=& -\sin t\,dt && v &=&\frac{1}{k}e^{kt} \end{array}

    And we have: . I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k}\left[\frac{1}{k}e^{kt}\cos t + \frac{1}{k}\int e^{kt}\sin t\,dt\right]

    . . . . . . . . . . I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k^2}e^{kt}\cos t - \frac{1}{k^2}\underbrace{\int e^{kt}\sin t\,dt}_{\text{This is }I} + C


    We have: . I + \frac{1}{k^2}I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k^2}e^{kt}\cos t + C

    Factor: . \frac{k^2+1}{k^2}\,I \;=\;\frac{e^{kt}}{k^2}(k\sin t -\cos t) + C

    Then: . I \;=\;\frac{k^2}{k^2+1}\cdot\frac{e^{kt}}{k^2}(k\si  n t - \cos t) + C

    Therefore: . I \;=\;\frac{e^{kt}}{k^2+1}(k\sin t - \cos t) + C


    because it was probably needed to solve a differential equation problem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 18th 2011, 01:12 AM
  2. Replies: 1
    Last Post: February 17th 2010, 03:58 PM
  3. How to find the integral of...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 10th 2010, 06:04 PM
  4. find the max value of integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 14th 2009, 05:19 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum