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?
Follow Math Help Forum on Facebook and Google+
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 $D_x(\arctan(x))=\dfrac{1}{1+x^2}~\&~D_x(\cot^{-1}(x)=\dfrac{-1}{1+x^2}~?$
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 ... $\arctan(x) = \dfrac{\pi}{2}-\text{arccot}(x)$
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 and . 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?
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