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JEE Main20209 Jan 2020Evening ShiftMathematicsIndefinite IntegrationActual

If ∫ d θ cos 2 θ tan 2 θ + sec 2 θ = λ tan θ + 2 log e f θ + C where C is a constant of integration, then the ordered pair λ , f θ is equal to:

Options

  1. A1 , 1 - tan θ
  2. B- 1 , 1 - tan θ
  3. C- 1 , 1 + tan θ
  4. D1 , 1 + tan θ

Correct answer

C. - 1 , 1 + tan θ

Step-by-step solution

∫ sec 2 θ 1 + tan 2 θ 1 - tan 2 θ + 2 tan θ 1 - tan 2 θ d θ = ∫ sec 2 θ 1 - tan 2 θ 1 + t a n θ 2 d θ = ∫ sec 2 θ 1 - tan θ 1 + tan θ d θ Let tan θ = t ⇒ sec 2 θ d θ = d t = ∫ 1 - t 1 + t d t = ∫ - 1 + 2 1 + t d t = - t + 2 ln 1 + t + C = - tan θ + 2 ln 1 + tan θ + C ⇒ λ = - 1   a n d   f θ = 1 + tan θ

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