Results 1 to 5 of 5

Math Help - Integration By Parts: What am I doing wrong?

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    20

    Question Integration By Parts: What am I doing wrong?

    \int 3xcos(5x)dx
    Let u = 3x so,  du = 3dx
    Let dv = cos(5x)dx so, v = sin(5x)/5

    Since, \int udv = uv - \int duv

    \int 3xcos(5x)dx = 3xsin(5x)/5 - (1/15)\int sin(5x)dx
     = 3xsin(5x)/5 - (1/15)[-cos(5x)/5]
     = (3/5)xsin(5x) + (1/75)cos(5x)

    ...but this is wrong.
    the answer is: (3/25)(cos(5x)+5xsin(5x))

    Where did I go wrong? Thank You
    Last edited by Krooger; February 3rd 2009 at 09:37 AM. Reason: Forgot brackets on real answer
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Krooger View Post
    \int 3xcos(5x)dx
    Let u = 3x so,  du = 3dx
    Let dv = cos(5x)dx so, v = sin(5x)/5

    Since, \int udv = uv - \int duv

    \int 3xcos(5x)dx = 3xsin(5x)/5 - ({\color{red}3/5})\int sin(5x)dx (You divided by 3 instead of multiplying by 3.)
     = 3xsin(5x)/5 - (1/15)[-cos(5x)/5]
     = (3/5)xsin(5x) + (1/75)cos(5x)

    ...but this is wrong.
    the answer is: (3/25)(cos(5x)+5xsin(5x))

    Where did I go wrong? Thank You
    ..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2009
    Posts
    20
    Ah I see, I was thinking you had to isolate dx in order to sub it in making my result du/3 instead of 3du.

    Thank You
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2009
    Posts
    26
    [quote=Krooger;259573] \int 3xcos(5x)dx
    Let u = 3x so,  du = 3dx
    Let dv = cos(5x)dx so, v = sin(5x)/5

    alright so u do uv-integral(vdu)
    so,
    3xsin(5x)/5-integral(sin(5x)/5*3dx
    3xsin(5x)/5+3cos(5x)/25
    do u understand how i got that?
    or need me to explain further?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2009
    Posts
    20
    Yup, got it I was thrown off from the fact that you can just use du, you dont have to isolate and replace dx. Because of that I was using du/3 not 3du.

    Thank You again
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration what am i doing wrong?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 19th 2011, 03:16 AM
  2. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  3. What did I do wrong? (integration)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 30th 2010, 05:07 AM
  4. What is wrong with my Integration?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 14th 2010, 01:21 PM
  5. Replies: 4
    Last Post: November 12th 2009, 11:41 AM

Search Tags


/mathhelpforum @mathhelpforum