If f ( x ) is the antiderivative of 1 + 2 tan x ( tan x + sec x ) 1 2 and f π 6 = l o g 2 , then the value of f ( 0 ) is

Question

If f ( x ) is the antiderivative of 1 + 2 tan x ( tan x + sec x ) 1 2 and f π 6 = l o g 2 , then the value of f ( 0 ) is,

MARKS App · Accepted Answer

f x = ∫ 1 + 2 tan ⁡ x tan ⁡ x + sec ⁡ x 1 2 d x ‍ = ∫ 1 + 2 tan 2 ⁡ x + 2 tan ⁡ x sec ⁡ x 1 2 d x ‍ = ∫ 1 + tan 2 ⁡ x + tan 2 ⁡ x + 2 tan ⁡ x sec ⁡ x 1 2 d x ‍ = ∫ sec 2 ⁡ x + tan 2 ⁡ x + 2 tan ⁡ x sec ⁡ x 1 2 d x ‍ = ∫ s e c x + t a n x 2 1 2 d x ‍ = log ⁡ s e c x + t a n x + log ⁡ sec ⁡ x + c = l o g s e c x s e c x + t a n x + c f π 6 = log 2 ⇒ log 2 3 2 3 + 1 3 + c = log 2 ⇒ c = 0 f x = log ⁡ sec ⁡ x sec ⁡ x + tan ⁡ x ⇒ f 0 = l o g 1 1 + 0 = 0

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