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∫ tan x dx =

Asin²x
Bsec²x
C-ln|cos x| + C (= ln|sec x| + C)
Dcos x
Answer & Solution
Correct answer: C. -ln|cos x| + C (= ln|sec x| + C)
∫ tan x dx = ∫ (sin x/cos x) dx. Let u = cos x, du = -sin x dx: = -∫ du/u = -ln|u| + C = -ln|cos x| + C = ln|sec x| + C.
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