Results 1 to 8 of 8

Math Help - integration on inverse trig

  1. #1
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106

    integration on inverse trig

     \int tan^{-1}u du

    is equal to

     utan^{-1}u-\frac{1}{2} ln(1+u^{2})

    can somebody explain to me how this is true?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    Quote Originally Posted by akhayoon View Post
     \int tan^{-1}u du is equal to
     utan^{-1}u-\frac{1}{2} ln(1+u^{2}) can somebody explain to me how this is true?
    Did you differentiate the answer?
    Do it!
    That shows you how it works.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106
    well thats the thing I haven't been taught these

    it's just given to me as a formula...but I would like to understand how it happened so that way I dont have to memorize it...

    thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    Quote Originally Posted by akhayoon View Post
    well thats the thing I haven't been taught these
    it's just given to me as a formula...but I would like to understand how it happened so that way I dont have to memorize it...
    Sorry to say, but I find that notion ridiculous.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106
    well..I really don't care you how you find that notion...all I really want is a hint on how to begin solving that problem really, which is what I always get when I have a question on this forum.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,901
    Thanks
    329
    Awards
    1
    Quote Originally Posted by akhayoon View Post
     \int tan^{-1}u du

    is equal to

     utan^{-1}u-\frac{1}{2} ln(1+u^{2})

    can somebody explain to me how this is true?
    Quote Originally Posted by Plato View Post
    Did you differentiate the answer?
    Do it!
    That shows you how it works.
    Here's your hint, as Plato suggested:
    \frac{d}{du} \left ( u~tan^{-1}(u)-\frac{1}{2} ln(1+u^{2}) \right )

    = tan^{-1}(u) + \frac{u}{u^2 + 1} - \frac{1}{2} \left ( \frac{1}{1 + u^2} \cdot 2u \right )

    What does this tell you?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    the responses given seem to me to be geared towards verifying that the claim is true, rather than saying why it's true, or how it works. i believe what the poster was after is actually integrating arctan to get the result shown. in that case, we use integration by parts with u = \arctan x and dv = 1 (you do know what i am referring to when i say u \mbox{ and } dv, right akhayoon?)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    A similar integration by parts reasoning is useful here: take the following function f(u)=u\arctan u. Contemplate its derivative and integrate, the rest follows.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse trig integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 5th 2010, 06:45 PM
  2. integration of inverse trig.
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 17th 2009, 05:23 AM
  3. more integration of inverse trig
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 4th 2009, 03:04 AM
  4. inverse trig integration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 10th 2009, 06:04 PM
  5. Inverse Trig Integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 10th 2008, 07:44 PM

Search Tags


/mathhelpforum @mathhelpforum