∫ 1 + tan 2 x 1 - tan 2 x d x =

Question

∫ 1 + tan 2 x 1 - tan 2 x d x =, log 1 - tan x 1 + tan x + c, log 1 + tan x 1 - tan x + c, 1 2 log 1 - tan x 1 + tan x + c, 1 2 log 1 + tan x 1 - tan x + c

MARKS App · Accepted Answer

Let, I = ∫ 1 + tan 2 x 1 - tan 2 x d x ⇒ I = ∫ sec 2 x 1 - tan 2 x d x Put tan x = t , we get sec 2 x d x = d t I = ∫ d t 1 - t 2 ⇒ I = 1 2 log 1 + t 1 - t + c ⇒ I = 1 2 log 1 + tan x 1 - tan x + c .

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