∫ sec²x dx =
Atan x + C
Bsin x
Cln|sec x|
Dcot x
Answer & Solution
Correct answer: A. tan x + C
d/dx(tan x) = sec²x, so ∫ sec²x dx = tan x + C. (And ∫ csc²x dx = -cot x + C, ∫ sec x tan x dx = sec x + C, ∫ csc x cot x dx = -csc x + C.)
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