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$\int \frac{\cos 2 x}{\cos x} d x is equal to$

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$2 \sin x + \log (\sec x - \tan x) + C$

$2 \sin x - \log (\sec x + \tan x) + C$

$2 \sin x + \log (\sec x + \tan x) + C$

$2 \sin x - \log (\sec x - \tan x) + C$

answer is D.

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Detailed Solution

$\begin{aligned} \int \frac{\cos 2 x}{\cos x} d x = \int 2 \cos x - \sec x d x \\ = 2 \sin x - \log (\sec x + \tan x) + C \end{aligned}$

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