how do i do this ? please show me with proper working
y= ln(ln(tan x))
also i did the one below , is it right ?
∫ sin y/cos^2 y= - cos y /cos^2 y= 1/cosy= sec y +c
thank you
Hello, arsenal12345!
How do I do this? .$\displaystyle y\:=\: \ln\left[\ln(\tan x)\right]$ . Do what?
I see no way to integrate it.
Do you want its derivative?
Also I did the one below ... is it right ?
. . $\displaystyle \int\frac{\sin y}{\cos^2\!y}\,dy \:=\:\underbrace{-\frac{\cos y}{\cos^2\!y}}_{{\color{red}??}} \:=\:\frac{1}{\cos y} \:=\:\sec y + C$
$\displaystyle \int\frac{\sin y}{\cos^2\!y}\,dy \;=\;\int\frac{1}{\cos y}\cdot\frac{\sin y}{\cos y}\,dy \;=\;\int\sec y\tan y\,dy \;=\;\sec y + C$