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Hi, hiw do we do this question?​

Hi, hiw do we do this question?​-example-1
User Parashuram
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1 Answer

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\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

User Aren Li
by
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