Results 1 to 6 of 6

Math Help - Arctangent Integration help

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    48

    Arctangent Integration help

    I need help integrating:

    xarctan(x) / ((1+x^2)^2)

    I did trig substitutions like this:

    u=tan(theta)
    du=sec^2(theta) d(theta)

    So I ended up with:

    Integral of (tan(theta)*arctan(tan(theta))*(sec^2(theta))) / ((1+tan^2(theta))^2)

    Which simplifies into:

    Integral of:

    (Theta)(tan(theta))(sec^2(theta)) / (sec^2(theta))^2

    The sec^2(theta)'s cancel so I have:

    (theta)(tan(theta))(cos^2(theta)), which simplifies into:

    (theta)(sin(theta))(cos(theta))

    Is that right, and if so, how do I integrate it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Try integration by parts with
    u=\tan^{-1}(x) and

    dv=\frac{x}{(x^2+1)^2}dx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
    48
    I also did that:

    u=arctanx
    du=1/(x^2+1)

    dv=x/(x^2+1)^2
    v= -1/(2(x^2+1))

    So then I get:

    -arctan(x)/(2(x^2+1)) + .5*Integral of 1/(1+x^2)^2

    That's UV - Integral of vdu. I don't know what to do from there...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    \int\frac{1}{(1+x^2)^2}dx

    set
    x=tan(\theta) then
    dx=sec^2(\theta)d\theta

    so you get

    \int\frac{1}{(1+\tan^2(\theta))^2}sec^2(\theta)d\t  heta=


    \int\frac{sec^2(\theta)}{sec^4(\theta)}d\theta=\in  t\cos^2(\theta)d\theta

    I think you can get it from here
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2008
    Posts
    48
    I took cos^2(theta) to (1/2)(1+cos(2theta)).

    So I now have it down to:

    -arctan(x)/(2(x^2+1)) + (1/2)*((1/2)theta + (1/4)sin (2theta))

    So, with the subsitution x=tan(theta), theta=arctan(x), so this factors to:

    (1/4)arctan(x) + (1/8)sin (2theta))

    What do I do with the (1/8)sin(2theta)? It's not good to put arctan(x) in the sin, to have (1/8)sin(2arctan(x)), right?

    I put it to 2sin(theta)cos(theta), so it's:

    (1/4)arctan(x) + (1/4)sin(theta)cos(theta)

    What's the best way to factor the last bit?
    Last edited by thegame189; February 28th 2008 at 01:08 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    since

    \tan(\theta)=\frac{x}{1}

    and by defintion

    tan(\theta)=\frac{opposite}{adjacent}

    we can figure the the hypotenuse using the pythagorean theorem...

    h=\sqrt{x^2+1}

    so using the defintion of trig functions...

    \sin(\theta)=\frac{opp}{hyp}=\frac{x}{\sqrt{x^2+1}  }

    by a similar argument

    \cos(\theta)=\frac{1}{\sqrt{x^2+1}}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: June 17th 2011, 10:17 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Arctangent Simplification
    Posted in the Geometry Forum
    Replies: 1
    Last Post: September 7th 2009, 03:08 PM
  4. improper arctangent integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 7th 2007, 06:28 PM
  5. Derive arctangent
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: August 1st 2007, 10:12 PM

Search Tags


/mathhelpforum @mathhelpforum