# Area under ArcTan[x]

1. Dec 22, 2003

### PrudensOptimus

Any errors? Please pick out and explain, thanks.

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

2. Dec 22, 2003

### kallazans

arctg(dy/dx)=tgx

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

3. Dec 22, 2003

### PrudensOptimus

Re: arctg(dy/dx)=tgx

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

4. Dec 23, 2003

### HallsofIvy

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

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

Last edited: Dec 23, 2003
5. Dec 23, 2003

### PrudensOptimus

ahh, what was i thinking:\

6. Dec 23, 2003

### PrudensOptimus

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

7. Jan 2, 2004

### sam2

Hi,

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

Sam

8. Jan 2, 2004

### PrudensOptimus

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

9. Jan 3, 2004

### 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.

10. Jan 3, 2004

### PrudensOptimus

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

11. Jan 4, 2004

### 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.