Integration: ∫ tan(x) dx =
Asec²x
Bsec x
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| = -ln|cos x| + C.
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