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AP EAMCET202023 Sep 2020Morning ShiftMathematicsIndefinite IntegrationActual

∫ 1 + tan 2 x 1 - tan 2 x d x =

Options

  1. Alog 1 - tan x 1 + tan x + c
  2. Blog 1 + tan x 1 - tan x + c
  3. C1 2 log 1 - tan x 1 + tan x + c
  4. D1 2 log 1 + tan x 1 - tan x + c

Correct answer

D. 1 2 log 1 + tan x 1 - tan x + c

Step-by-step solution

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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