A) \[2\sin x+\log (\sec x-\tan x)+c\]
B) \[2\sin x-\log (\sec x-\tan x)+c\]
C) \[2\sin x+\log (\sec x+\tan x)+c\]
D) \[2\sin -\log (\sec x+\tan x)+c\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\frac{\cos 2xdx}{\cos x}}\] \[=\int_{{}}^{{}}{\frac{2{{\cos }^{2}}x-1}{\cos x}}=\int_{{}}^{{}}{(2\cos x-\sec x)}dx\] \[=2\sin x=\log (\sec x+\tan x)+c\]
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