Area under ArcTan[x]

[PrudensOptimus]

Any errors? Please pick out and explain, thanks.

\int{}tan^{-1}(x)dx = F(x)
F'(x) = tan^{-1}(x)
\frac{dy}{dx} = tan^{-1}(x)
dy = tan^{-1}(x) dx
tan^{-1}(\frac{dy}{dx}) = tan(x)
\frac{F'(x)}{1+F^{2}(x)} = sec^{2}(x)
F'(x) = sec^{2}(x)[1 + F^{2}(x)]
F(x) = tan(x) + \int{}\frac{sin(x)}{cos^{3}(x)}dx
F(x) = tan(x) + \frac{1}{2cos^{2}(x)} + C

[kallazans]

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



your question is very interesting!

[PrudensOptimus]

Originally posted by kallazans
can you show me arctg(dy/dx)=tgx?



your question is very interesting!


I think I already typed out all my steps...

[HallsofIvy]

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

\frac{dy}{dx}= tan^{-1}(x) so
tan(\frac{dy}{dx})= x

[PrudensOptimus]

Originally posted by HallsofIvy
This step is wrong:
dy = tan^{-1}(x) dx
tan^{-1}(\frac{dy}{dx}) = tan(x)

\frac{dy}{dx}= tan^{-1}(x) so
tan(\frac{dy}{dx})= x

ahh, what was i thinking:\

[PrudensOptimus]

Any other way to find ArcTan[x] area? No Parts please.

[sam2]

Hi,

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

Sam

[PrudensOptimus]

Originally posted by sam2
Hi,

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

Sam

Well, sometimes the other way might define a new method of solving harder problems.

[Sonty]

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.

[PrudensOptimus]

Originally posted by Sonty
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.


Isn't Tan(x) and Tan^-1(X) a MacLauren series?

[Sonty]

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.