Question

Hi, hiw do we do this question?​

Answer 1

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

Explanation:

`\displaystyle \int\sec x\:dx=\int\sec x\left((\sec x+ \tan x)/(\sec x + \tan x)\right)dx`

`\displaystyle = \int \left((\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((\sec x\tan x + \sec^2x)/(\sec x + \tan x)\right)\:dx = \int (du)/(u)`

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