How do you integrate $secx$ ?

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Answer 1 · 2016-09-02T16:33:57

The standard trick is to multiply by $(secx+tanx)/(secx+tanx)$ , then use substitution.

$int sec x dx = int secx (secx+tanx)/(secx+tanx) dx$

$= int (sec^2x+secxtanx)/(secx+tanx) dx$

With $u = secx+tanx$ , we have $int 1/u du$ , so

$int sec x dx = ln abs(sec x + tan x) +C$

How do you integrate secxsecx?