Results 1 to 6 of 6

Math Help - integration by parts

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    106

    integration by parts

    i know this probably seems really trivial, but in college calc we basically have to teach ourselves.. and i don't understand where i'm going wrong >.<

    the problem is, integral (e^x)(cosx)dx

    so i figured this is something you'd go about by using parts

    u= e^x v= sinx
    du= e^xdx dv= cosxdx

    then you get e^x(sinx)- integral e^x(sinx)

    doing parts again.. u= sin x v= e^x
    du= cosxdx dv= e^xdx

    and it seems like i'm going to get e^x(sinx)-e^x(sinx) * integral (e^x)(cosx)dx which is where i started..

    this is a mistake that i often make in parts.. please help? i need to be able to understand/explain what i did
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2009
    Posts
    66
    Quote Originally Posted by buttonbear View Post
    i know this probably seems really trivial, but in college calc we basically have to teach ourselves.. and i don't understand where i'm going wrong >.<

    the problem is, integral (e^x)(cosx)dx

    so i figured this is something you'd go about by using parts

    u= e^x v= sinx
    du= e^xdx dv= cosxdx

    then you get e^x(sinx)- integral e^x(sinx)

    doing parts again.. u= sin x v= e^x
    du= cosxdx dv= e^xdx

    and it seems like i'm going to get e^x(sinx)-e^x(sinx) * integral (e^x)(cosx)dx which is where i started..

    this is a mistake that i often make in parts.. please help? i need to be able to understand/explain what i did
    I don't think integration by parts is the way to go for this. As you say, it doesn't make it any easier.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2009
    Posts
    106
    okay..

    can you give me a hint?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2009
    Posts
    66
    Quote Originally Posted by buttonbear View Post
    okay..

    can you give me a hint?
    Oh I'd love to, I'm thinking about it myself
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,677
    Thanks
    446
    Quote Originally Posted by buttonbear View Post
    i know this probably seems really trivial, but in college calc we basically have to teach ourselves.. and i don't understand where i'm going wrong >.<

    the problem is, integral (e^x)(cosx)dx

    so i figured this is something you'd go about by using parts

    u= e^x v= sinx
    du= e^xdx dv= cosxdx

    then you get e^x(sinx)- integral e^x(sinx)

    doing parts again.. u= sin x v= e^x
    du= cosxdx dv= e^xdx

    and it seems like i'm going to get e^x(sinx)-e^x(sinx) * integral (e^x)(cosx)dx which is where i started..

    this is a mistake that i often make in parts.. please help? i need to be able to understand/explain what i did
    \int e^x \cos{x} \, dx

    u = \cos{x} ... dv = e^x dx

    du = -\sin{x} \, dx ... v = e^x

    \int e^x \cos{x} \, dx = e^x\cos{x} + \int e^x \sin{x} \, dx

    u = \sin{x} ... dv = e^x \, dx

    du = \cos{x} \, dx ... v = e^x

    \int e^x \cos{x} \, dx = e^x\cos{x} + e^x \sin{x} - \int e^x \cos{x} \, dx

    2\int e^x \cos{x} \, dx = e^x\cos{x} + e^x \sin{x}

    \int e^x \cos{x} \, dx = \frac{e^x(\cos{x} + \sin{x})}{2} + C
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2009
    Posts
    106
    thanks a lot for your help

    so, basically i was just doing the substitutions backwards?

    i guess i should just try to be more persistent and not give up so quickly..
    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