$\displaystyle \int{\sqrt{sec^2x}} $
The square root is giving me some problems here. I know that the identity of $\displaystyle sec^2 x$ is $\displaystyle tan x$, but how does the sqrt play here?
the square root cancels the square. (technically you would have to use the absolute value, but it usually doesn't cause problems if we forget it, especially for indefinite integrals).
so you want $\displaystyle \int \sec x~dx$
now there's an interesting integral ...if you don't know it by heart, that is
the identity is $\displaystyle \sec^2 x = 1 + \tan^2x$. but you don't need that hereThe square root is giving me some problems here. I know that the identity of $\displaystyle sec^2 x$ is $\displaystyle tan x$, but how does the sqrt play here?