Results 1 to 6 of 6

Math Help - Integration of nasty trig product

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    32

    Integration of nasty trig product

    Hi all,

    Would anybody be able to explain the technique for integrating something like this:

    4p^2sin(10p^3-5) dp

    I've been trying to do it by parts but with no success at all.
    Any help appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by DangerousDave View Post
    Hi all,

    Would anybody be able to explain the technique for integrating something like this:

    4p^2sin(10p^3-5) dp

    I've been trying to do it by parts but with no success at all.
    Any help appreciated.
    Don't use integration by parts. Use the substitution u = 10p^3 - 5.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    32
    Cheers for response - but it still seems to be turning out pretty hideous. Am I to reformulate 4p^2 in terms of u too (and of course get du in terms of dp) to get a full expression in terms of u to integrate with respect to u? It seems like that results in a simlarly nasty-looking product in need of integration.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by DangerousDave View Post
    Cheers for response - but it still seems to be turning out pretty hideous. Am I to reformulate 4p^2 in terms of u too (and of course get du in terms of dp) to get a full expression in terms of u to integrate with respect to u? It seems like that results in a simlarly nasty-looking product in need of integration.
    \int 4 p^2 \sin (10 p^3 - 5) \, dp becomes \frac{2}{15} \int \sin (u) \, du.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    Posts
    32
    I think now I'm beginning to discover exactly where one of the main missing chunks in my education in calculus lies. I think I understand what's going on here:

    du/dp = 30p^2, therefore dp = du/30p^2

    that cancels immediately with 4P^2 to produce 2/15, which can then be removed in front of the integral.

    My weakness is that I never really got to grips with the significance of the d-terms, and what it was permissible to do with them. 'With respect to' is such a non-mathematical sounding idea, that I didn't really know what to do with 'with respect to u divided by 30p^2' - that seems like comparing apples and oranges to me, and hard to see as a seamless part of the mathematical expression. Oh well, thanks for the help anyway.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by DangerousDave View Post
    I think now I'm beginning to discover exactly where one of the main missing chunks in my education in calculus lies. I think I understand what's going on here:

    du/dp = 30p^2, therefore dp = du/30p^2

    that cancels immediately with 4P^2 to produce 2/15, which can then be removed in front of the integral.

    My weakness is that I never really got to grips with the significance of the d-terms, and what it was permissible to do with them. 'With respect to' is such a non-mathematical sounding idea, that I didn't really know what to do with 'with respect to u divided by 30p^2' - that seems like comparing apples and oranges to me, and hard to see as a seamless part of the mathematical expression. Oh well, thanks for the help anyway.
    You seem to have the right idea now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Nasty limit...
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 14th 2011, 09:44 PM
  2. Trig product
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 16th 2009, 08:30 AM
  3. A Nasty integral -
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 22nd 2008, 07:16 PM
  4. nasty trig induction
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: August 29th 2006, 05:29 AM
  5. A nasty integration
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 23rd 2006, 03:53 AM

Search Tags


/mathhelpforum @mathhelpforum