Results 1 to 2 of 2

Math Help - integration by parts.

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    UK
    Posts
    16

    integration by parts.

    can someone confirm if i am doing this correctly

    5xcos(4x) solve using integration by parts


    let V=5x so dv/dx=5

    du/dx=cos(4x) u=1/4sin4x

    5x(1/4sin4x)-int 5(1/4sin4x)

    (5x/4)sin4x + (5/16)cos4x

    if i am going wrong somewhere can you please explain thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: integration by parts.

    It looks to me that you have correctly applied the method, although don't forget the constant of integration.

    If you have doubts about your indefinite integration result, recall that you may use differentiation as a check:

    \frac{d}{dx}\left(\frac{5x}{4}\sin(4x)+\frac{5}{16  }\cos(4x)+C\right)=\frac{5x}{4}(4\cos(4x))+\frac{5  }{4}\sin(4x)-\frac{5}{4}\sin(4x)+0=5x\cos(4x)

    Since the derivative of the anti-derivative is the original integrand, you know the result is correct.
    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: 2
    Last Post: April 7th 2009, 03:28 PM
  5. Replies: 1
    Last Post: February 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum