# I Equivalent form for arctan?

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1. Feb 17, 2017

### Aristotle

Hi, I was just looking at an example for a certain problem and noticed that in the second step they went to arctan(epsilon). I know there's a form that is equal to arctan but am a little unsure.
I've come across formulas on the web such as
arctan(x) = ∫(dt)/(a2+t2)
but nothing else that would get to arctan.

Can somebody please direct me to the correct formula?

2. Feb 17, 2017

### Staff: Mentor

3. Feb 17, 2017

### Staff: Mentor

I think the $\varepsilon$ in the term with $\tan^{-1}$ is misleading. What if you substitute $u := \frac{t}{\varepsilon}$ and solve the integral?

4. Feb 17, 2017

### Aristotle

If you don't mind me asking, where did you get t/ε from?

5. Feb 17, 2017

### Staff: Mentor

For the integral, $\varepsilon$ is only a disturbing constant as $\pi$ is. I simply tried to get the integral in the form $\int \frac{1}{1+x^2} dx$ which means to pull out $\varepsilon^2$ in the denominator.

6. Feb 17, 2017

### Aristotle

The only way I see taking ε2 out of the denominator is dividing that number by itself for numerator and denominator.
But you would get ∫(1/ε)⋅( (dt) / ( (t22)+1) )

7. Feb 17, 2017

### Staff: Mentor

Yes, and now substitute $u := \frac{t}{\varepsilon}$ and replace $dt$ by $du$.

8. Feb 17, 2017

### Aristotle

Wouldn't that get you arctan(t) and not arctan(ε)?

9. Feb 17, 2017

### Staff: Mentor

$\arctan u$ as I see it. I don't understand how the $\varepsilon$ get's into the argument of $\arctan$. As such it is the variable where the $\pm \infty$ apply to, not the $\varepsilon$ from the limit. I said I find the notation misleading. But it's only a temporary result anyway, so I didn't bother too much.

10. Feb 17, 2017

### Aristotle

Thank you so much for your help! I also got the same. Possibly their answer is incorrect...

11. Feb 17, 2017

### Staff: Mentor

Why? It's $1$ in the end, so the result is correct, only the $\varepsilon$ in between is odd.

12. Feb 17, 2017

### Aristotle

Woops didnt mean to say the answer was wrong.
But yeah the ε in the arctan in that step is odd. Think they forgot a 't' in the numerator.