Results 1 to 3 of 3

Math Help - Integration by Parts: Using 1*dx as your dv value

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    8

    Integration by Parts: Using 1*dx as your dv value

    Not sure if you guys use a different formula (maybe different representation of variables) so I"ll put down my I B P formula:

    (Integral of) u*dv = u*v - (Integral of) v*du

    Now i've been looking at a couple of examples and in most of them (especially where you are going to end up with a dv value very hard to integrate) u is the entire term (not including dx of course) and dv is just 1*dx. Is this a very common/effective method?

    P.S. I'm new to this forum/forums in general, how do you get the fancy math symbols on here
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by someguy456345 View Post
    Now i've been looking at a couple of examples and in most of them (especially where you are going to end up with a dv value very hard to integrate) u is the entire term (not including dx of course) and dv is just 1*dx. Is this a very common/effective method?
    It really depends on the integrand. But yes, sometimes this is the way to go.

    There is a nice rule of thumb for choosing u known as the LIATE rule (though its more a guideline than a rule; there are always exceptions). Among several different types of factors, choose u to be the factor that comes first in this list.

    Logarithmic functions
    Inverse trigonometric functions
    Algebraic functions
    Trigonometric functions
    Exponential functions

    For example, when doing \mbox{$\int\arcsin x\,dx$}, we have an inverse trigonometric factor, \arcsin x, and an algebraic factor, 1. Using the above guideline, we let u=\arcsin x,\;dv=dx.

    But if you were doing \mbox{$\int xe^x\,dx$}, you would prefer the algebraic factor for u, leaving the exponential for dv. So, u=x,\;dv=e^x\,dx.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    8
    Wow thank you so much, this is quite possibly one of the best things someone has ever shown me in 5 minutes. The LIATE rule makes perfect sense to me, as in my eyes anyway, the rule goes from hardest to easiest to integrate for subbing in U.

    I really appreciate the response, thanks!
    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