- #1

karush

Gold Member

MHB

- 3,269

- 5

$\tiny{213(DOY)}$

$\displaystyle\int \sec x \tan x \: dx =$

(A) $\sec x + C$

(B) $\tan x + C$

(C) $\dfrac{\sec^2 x}{2}+ C$

(D) $\dfrac {tan^2 x}{2}+C $

(E) $\dfrac{\sec^2 x \tan^2 x }{2}+ C$

$\displaystyle\int \sec x \tan x \: dx =$

(A) $\sec x + C$

(B) $\tan x + C$

(C) $\dfrac{\sec^2 x}{2}+ C$

(D) $\dfrac {tan^2 x}{2}+C $

(E) $\dfrac{\sec^2 x \tan^2 x }{2}+ C$

$(\sec x)'=\sec x \tan x$ so the answer is $\sec x +C$ (A)

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