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Integration of x(arctanx)^2

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data

    I have to integrate using the following integration methods:
    1) u-sub
    2) by parts
    3) trig
    4) partial fraction



    2. Relevant equations

    [tex]\int[/tex]x(arctanx)2

    3. The attempt at a solution

    According to me, the method I should use is integration by parts.

    When I tried it, the whole thing got worse...

    u = arctanx2
    du = 2arctanx/1+x2
    dv = xdx
    v = 1/2x2

    [tex]\int[/tex]x(arctanx)2 = 1/2x2(arctanx)2 - [tex]\int[/tex]x2arctanx/1+x2
     
  2. jcsd
  3. Feb 22, 2010 #2
    I would use integration by parts, though you're going to need to use it twice.

    First use IBP to find calculate [itex] \int \arctan(x) dx [/itex]. Once you have this, you can use IBP on the whole thing.
     
  4. Feb 22, 2010 #3
    try using a U substitution, say u=arctan(x) and [tex]du=\frac{1}{1+x^{2}}dx[/tex]
    then dx=(1+x[tex]^{2}[/tex])du , where x=tan(u). also plugging in for the x multiplied in front gets you [tex]\int u^{2}tan(u)(1+tan(u)^{2})du[/tex]
     
    Last edited: Feb 22, 2010
  5. Feb 22, 2010 #4
    That is definitely more elegant, though I'm not sure that Ayesh will see the trick involved to make it an easy IBP.
     
  6. Feb 23, 2010 #5
    Kreizhn why should I use [tex]\int[/tex]arctanx first?
     
  7. Feb 23, 2010 #6
    I get from where dx(1 + tan(u)2)du comes from, but not the u2tan(u).
     
  8. Feb 23, 2010 #7
    If you're going to keep going in the more elegant direction, I wouldn't worry about integrating arctan first. It is a method that works, but takes a lot more work.

    As for your other question, consider that your integrand is [itex] x(\arctan(x))^2 [/itex] and you make the substitution [itex] u = \arctan(x) [/itex] or alternatively, [itex] x = \tan(u) [/itex]. Now when making the substitution, there are three things you are going to need to account for, namely, the x, the [itex] \arctan^2(x) [/itex] and the dx.

    see if you can follow the following bit of arithmetic

    [tex] \begin{align*}
    x \arctan^2(x) dx &= x u^2 dx & \text{ since } u=\arctan(x) \\
    &= tan(u) u^2 dx & \text{ since } x = \tan(u) \\
    &= u^2 \tan(u) (1+x^2) du & \text{ since } dx = (1+x^2) du \\
    &= u^2 \tan(u) (1+tan^2(u) )du & \text{ since } x = \tan(u)
    \end{align*}
    [/tex]

    The tricky part now is realizing that there is an implicit derivative in here that will make integration by parts simple.
     
  9. Feb 24, 2010 #8
    Thank you!

    Your explanations are very clear!
     
  10. Feb 24, 2010 #9
    This is what I have done until now:

    [tex]\int[/tex]x(arctanx)^2 dx

    1/2 x2arctanx - integral 1/2x2(1/1+x^2) dx

    1/2x2arctanx - 1/2 integral tan2t/1+tan2t *sec2t

    1/2x2arctanx - 1/2 integral tan2t

    1/2x2arctanx - 1/2 integral (sec2t - 1) dt

    1/2x2arctanx - 1/2 integral (tant - t) dt

    ... ?

    I don't what to do after.
    I know it has something to do with drawing a triangle, but I don't know what values to put around it.
     
    Last edited: Feb 24, 2010
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