Question

Hi, hiw do we do this question?​

Answer 1

$\displaystyle \int\sec x\:dx = \ln |\sec x + \tan x| + C$

Step-by-step explanation:

$\displaystyle \int\sec x\:dx=\int\sec x\left(\frac{\sec x+ \tan x}{\sec x + \tan x}\right)dx$

$\displaystyle = \int \left(\dfrac{\sec x\tan x + \sec^2x}{\sec x + \tan x} \right)dx$

Let $u = \sec x + \tan x$

$\:\:\:\:\:\:du = (\sec x\tan x + \sec^2x)dx$

where

$d(\sec x) = \sec x\tan x\:dx$

$d(\tan x) = \sec^2x\:dx$

$\displaystyle \Rightarrow \int \left(\frac{\sec x\tan x + \sec^2x}{\sec x + \tan x}\right)\:dx = \int \dfrac{du}{u}$

$= \ln |u| + C = \ln |\sec x + \tan x| + C$