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14 mark for review $int sec x \tan x dx =$ a $sec x + c$ b $\tan x + c$…

Question

14 mark for review
$int sec x \tan x dx =$
a $sec x + c$
b $\tan x + c$
c $\frac{sec^{2}x}{2} + c$
d $\frac{\tan^{2}x}{2} + c$
e $\frac{sec^{2}x\tan^{2}x}{2} + c$

Explanation:

Step1: Recall derivative of sec x

The derivative of $\sec x$ is $\sec x \tan x$, i.e., $\frac{d}{dx}(\sec x) = \sec x \tan x$.

Step2: Apply antiderivative definition

Since integration is the reverse of differentiation, the antiderivative of $\sec x \tan x$ is $\sec x + C$ (where $C$ is the constant of integration).

Answer:

A. $\sec x + C$