Q. $\int sin 2 x d (tan x)$ is equal to

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A

$2 lo g ∣ cos x ∣ + C$

B

$lo g ∣ cos x ∣ + C$

C

$2 lo g ∣ sec x ∣ + C$

D

$lo g ∣ sec x ∣ + C$

Solution:

$I = \int sin 2 x d (tan x) = \int sin 2 x . sec^{2} x d x$ $= 2 \int tan x d x = 2 lo g ∣ sec x ∣ + C$