Is there an equivalent form for arctan?

[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?

 

[jedishrfu]

This article has many variants for all the arc functions

https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#arctan

 

[fresh_42]

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

 

[Aristotle]

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

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

 

[fresh_42]

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.

 

[Aristotle]

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.

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) / ( (t²/ε²)+1) )

 

[fresh_42]

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

 

[Aristotle]

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

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

 

[fresh_42]

$\arctan u$ 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.

 

[Aristotle]

$\arctan u$ 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.

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

 

[fresh_42]

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

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

 

[Aristotle]

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

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.