How do you integrate $sec x$ $secx$ ?

Question

3753 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-02T16:33:57

The standard trick is to multiply by $\frac{sec x + tan x}{sec x + tan x}$ $(secx+tanx)/(secx+tanx)$ , then use substitution.

$\int sec x d x = \int sec x \frac{sec x + tan x}{sec x + tan x} d x$ $int sec x dx = int secx (secx+tanx)/(secx+tanx) dx$

$= \int \frac{{sec}^{2} x + sec x tan x}{sec x + tan x} d x$ $= int (sec^2x+secxtanx)/(secx+tanx) dx$

With $u = sec x + tan x$ $u = secx+tanx$ , we have $\int \frac{1}{u} d u$ $int 1/u du$ , so

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

How do you integrate secxsecx?