1. ## Simple Integral

Integration of 1/(1 + x^2)

when I use x = tan x, I get the answer as tan^-1(x)

when I use x = cot x, I get the answer as -cot^-1(x)

Does this mean tan^-1(x) = -cot^-1(x)? How is this possible?

2. ## Re: Simple Integral

Originally Posted by joshuaa
Integration of 1/(1 + x^2)

when I use x = tan x, I get the answer as tan^-1(x)

when I use x = cot x, I get the answer as -cot^-1(x)

Does this mean tan^-1(x) = -cot^-1(x)? How is this possible?
Don't you know that $D_x(\arctan(x))=\dfrac{1}{1+x^2}~\&~D_x(\cot^{-1}(x)=\dfrac{-1}{1+x^2}~?$

3. ## Re: Simple Integral

Originally Posted by joshuaa
Integration of 1/(1 + x^2)

when I use x = tan x, I get the answer as tan^-1(x)

when I use x = cot x, I get the answer as -cot^-1(x)

Does this mean tan^-1(x) = -cot^-1(x)? How is this possible?
$\arctan(x)$ and $-\text{arccot}(x)$ differ by a constant ...

$\arctan(x) = \dfrac{\pi}{2}-\text{arccot}(x)$

4. ## Re: Simple Integral

Originally Posted by joshuaa
Integration of 1/(1 + x^2)

when I use x = tan x, I get the answer as tan^-1(x)

when I use x = cot x, I get the answer as -cot^-1(x)
You mean, I hope, "u= tan(x)" and "u= cot(x)". And you should have got $tan^{-1}(x)+ C$ and $cot^{-1}(x)+ C$. As others have said, yes, those are the same with different constants.

Does this mean tan^-1(x) = -cot^-1(x)? How is this possible?

5. ## Re: Simple Integral

Originally Posted by Plato
Don't you know that $D_x(\arctan(x))=\dfrac{1}{1+x^2}~\&~D_x(\cot^{-1}(x)=\dfrac{-1}{1+x^2}~?$
Just realized now.

thanks guys