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Area under ArcTan[x]

  1. Dec 22, 2003 #1
    Any errors? Please pick out and explain, thanks.

    [tex]\int{}tan^{-1}(x)dx = F(x)[/tex]
    [tex]F'(x) = tan^{-1}(x)[/tex]
    [tex]\frac{dy}{dx} = tan^{-1}(x)[/tex]
    [tex]dy = tan^{-1}(x) dx[/tex]
    [tex]tan^{-1}(\frac{dy}{dx}) = tan(x) [/tex]
    [tex]\frac{F'(x)}{1+F^{2}(x)} = sec^{2}(x)[/tex]
    [tex]F'(x) = sec^{2}(x)[1 + F^{2}(x)][/tex]
    [tex]F(x) = tan(x) + \int{}\frac{sin(x)}{cos^{3}(x)}dx[/tex]
    [tex]F(x) = tan(x) + \frac{1}{2cos^{2}(x)} + C[/tex]
  2. jcsd
  3. Dec 22, 2003 #2

    can you show me arctg(dy/dx)=tgx?

    your question is very interesting!
  4. Dec 22, 2003 #3
    Re: arctg(dy/dx)=tgx

    I think I already typed out all my steps...
  5. Dec 23, 2003 #4


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    Science Advisor

    This step is wrong:
    [tex]dy = tan^{-1}(x) dx[/tex]
    [tex]tan^{-1}(\frac{dy}{dx}) = tan(x) [/tex]

    [tex]\frac{dy}{dx}= tan^{-1}(x)[/tex] so
    [tex]tan(\frac{dy}{dx})= x[/tex]
    Last edited by a moderator: Dec 23, 2003
  6. Dec 23, 2003 #5
    ahh, what was i thinking:\
  7. Dec 23, 2003 #6
    Any other way to find ArcTan[x] area? No Parts please.
  8. Jan 2, 2004 #7

    Why not using integration by parts? It's easy to do it that way.

  9. Jan 2, 2004 #8
    Well, sometimes the other way might define a new method of solving harder problems.
  10. Jan 3, 2004 #9
    I guess you could expand it in a series and integrate each term, then pick up the pieces again. If the series wouldn't turn out to be infinite as in this case all will be swell.
  11. Jan 3, 2004 #10

    Isn't Tan(x) and Tan^-1(X) a MacLauren series?
  12. Jan 4, 2004 #11
    you can expand around 0, of course, or around any other point. The annoying thing is that in the end you have to make all those convergence calculations. you can even go into a fourier expansion so you won't be integrating polynomials, but cos and sin. whatever. you can always find harder ways to solve simple problems.
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