Results 1 to 15 of 15

Math Help - Simpl- integral

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    541

    Simpl- integral

    Calculate :
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by apcalculus View Post
    you didn't post anything

    Quote Originally Posted by dhiab View Post
    Calculate : \int_0^1 \frac {x^3}{x^6 + 2x^3 + 1}~dx
    it seems like a pain, but i don't see an easier way to approach this at the moment. here's what i'd do

    write as \int_0^1 \frac {x^3}{(x^3 + 1)^2}~dx = \int_0^1 \frac {x^3}{(x + 1)^2(x^2 - x + 1)^2}~dx

    now begin the painful method of partial fractions
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    228
    What? It can be easily factored as (x^3 + 1)^2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Kaitosan View Post
    What? It can be easily factored as (x^3 + 1)^2.
    that has already been noted. are you saying you see an easier way to proceed than partial fractions?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    228
    Maybe trigonometric substitution would be easier, I don't know.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Kaitosan View Post
    Maybe trigonometric substitution would be easier, I don't know.
    trig substitution requires one of the forms: x^2 + a^2, x^2 - a^2, or a^2 - x^2 for some nonzero constant a. none of those forms show up here
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2008
    Posts
    228
    Wow. I really do think that's wrong. Trig sub doesn't have that form. It has the form of u^2 + a^2 and similar ones in which "u" is an expression that involves a variable like "x".
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,415
    Thanks
    1853
    My fault, I misread this completely!
    Last edited by HallsofIvy; June 21st 2009 at 04:42 AM. Reason: Added the closing latex tag
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Dec 2008
    Posts
    228
    I'm confused. What point did I make? lol

    I'm not going out to do all the math but here's the bottom line -

    In some cases, either partial fractions as well as trig sub can be used. I believe this is one of those cases in which u = x ^ (3/2). Then you solve the eqution u=a(one trig expression) isolating x.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by HallsofIvy View Post
    No, (x+1)^2 is NOT equal to [tex](x+1)^2(x^2- x+ 1). That was Kaitosan's point!
    What Jhevon wrote is correct... What you copied is ... a nonsense ?

    @ Kaitosan : you need to factorise into (irreducible) polynomials with smaller degree
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5

    Mr F's case study in Diplomacy and Tact

    Quote Originally Posted by HallsofIvy View Post
    No, (x+1)^2 is NOT equal to [tex](x+1)^2(x^2- x+ 1). That was Kaitosan's point!
    Since it's a little known fact that moo is studying Diplomacy and Tact 101 and failing badly (the former is the little known fact, by the way), I will jump in here using HallsofIvy's post as a case study:

    *Ahem* I think Jhevon said that (x^3 + 1)^2 = (x + 1)^2 (x^2 - x + 1)^2, which is a correct identity. His point in saying this was to get the linear and quadratic factors required for the denominators of the partial fraction decomposition.

    Kaitosan's point (and just because you don't think you made one doesn't mean you didn't ) was that a trignometric substitution can be made in a wider range of forms than Jhevon initially listed. Note: For some of these forms a hyperbolic substitution probably provides an easier path.


    I'm thinking of using this example in my next book: Diplomacy and Tact for Bovines. (There will be a French edition). (And there will be an Appendix on latex tags .... )

    You may wish to view my reply in this thread for a further example: http://www.mathhelpforum.com/math-he...tml#post331889
    Last edited by mr fantastic; June 21st 2009 at 02:08 AM. Reason: Corrected typo - thanks moo.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Haha !

    No, in this case, the problem is that I didn't understand what HallsofIvy wanted to say, as I misread "Kaitosan's point" and with the typos, it's not easier...

    *Ahem* I think Jhevon said that (x^3 + 1)^2 = (x + 1)^2 (x^2 - x + 1)^{\color{red}2}, which is a correct identity.
    Better this way
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by dhiab View Post
    Calculate :
    put \frac{3x^2}{(x^3+1)^2} \ dx = dv, \ \frac{1}{3}x=u. then v=\frac{-1}{x^3+1} and du=\frac{1}{3}dx. applying integration by parts will give you \int_0^1 \frac{x^3}{(x^3 + 1)^2} \ dx = \frac{-1}{6} + \frac{1}{3}\int_0^1 \frac{dx}{x^3 + 1}. i guess you can finish the job now.

    the final answer, according to my calculation, is \frac{\pi \sqrt{3}}{27} + \frac{1}{9} \ln 2 - \frac{1}{6}.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member
    Joined
    Jan 2009
    Posts
    715
    If the upper limit of the integral is infinity , it will be very well
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Dec 2008
    Posts
    228
    Quote Originally Posted by Moo View Post
    What Jhevon wrote is correct... What you copied is ... a nonsense ?

    @ Kaitosan : you need to factorise into (irreducible) polynomials with smaller degree
    That's only if I want to use partial fraction integration!!!! I can use trig sub as well if I want.

    Edit: Aah, I see. Trig sub is only valid if you can solve the integral of (tanx)^(2/3) lol
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simpl-integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 28th 2010, 12:22 PM
  2. Simpl-equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 31st 2009, 04:15 AM
  3. Very simpl integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 17th 2009, 02:10 PM
  4. simpl inequation
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: July 16th 2009, 11:24 AM
  5. Simpl-equation in N
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: June 20th 2009, 01:22 AM

Search Tags


/mathhelpforum @mathhelpforum