Results 1 to 4 of 4

Math Help - Integration by Parts

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    48

    Integration by Parts

    Hi, can anyone help me solve this:

    First make a substitution then use integration by parts to evaluate the integral:

    \int^\pi_0 e^{\cos{t}}\sin{2t} dt

    So far I have:

    \int^\pi_0 e^{\cos{t}}2\sin{t}\cos{t} dt

    let u = \cos{t}

    How do I do the next part?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,688
    Thanks
    447
    Quote Originally Posted by coldfire View Post
    Hi, can anyone help me solve this:

    First make a substitution then use integration by parts to evaluate the integral:

    \int^\pi_0 e^{\cos{t}}\sin{2t} dt

    So far I have:

    \int^\pi_0 e^{\cos{t}}2\sin{t}\cos{t} \, dt

    let u = \cos{t}

    How do I do the next part?
    du = -\sin{t} \, dt

    -2 \int^\pi_0 e^{\cos{t}}(-\sin{t})\cos{t} dt

    -2 \int_1^{-1} e^u \cdot u \, du

    2 \int_{-1}^{1} e^u \cdot u \, du

    now do parts
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by coldfire View Post
    Hi, can anyone help me solve this:


    First make a substitution then use integration by parts to evaluate the integral:

    \int^\pi_0 e^{\cos{t}}\sin{2t} dt

    So far I have:

    \int^\pi_0 e^{\cos{t}}2\sin{t}\cos{t} dt

    let u = \cos{t}

    How do I do the next part?
    Spoiler:

    Then du = -\sin t \ dt. Your integral went from t=0 to t=\pi; now it should go from u = \cos 0 = 1 to u = \cos \pi = -1.

    so

    I = -2\int_1^{-1} e^{u}u\ du

    Now we use integration by parts with dv = e^u, w=u. Then

    I = -2\left((ue^u)^{u=-1}_{u=1}-\int_1^{-1} e^{u}\ du\right)

    =-2\left((-e^{-1}-e)-(e^{-1}-e)\right)

    =4e^{-1}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2008
    Posts
    48
    Thanks guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2012, 02:30 PM
  2. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  3. Replies: 0
    Last Post: April 23rd 2010, 03:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 22nd 2010, 03:19 AM
  5. Replies: 1
    Last Post: February 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum