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JEE Main20199 Apr 2019Evening ShiftMathematicsIndefinite IntegrationActual

If ∫ e s e c x s e c x tan x f x + s e c x tan x + s e c 2 x d x = e s e c x f x + C , then a possible choice of f x is:

Options

  1. As e c x - t a n x - 1 2
  2. Bs e c x + t a n x + 1 2
  3. Cx s e c x + t a n x + 1 2
  4. Ds e c x + x t a n x - 1 2

Correct answer

B. s e c x + t a n x + 1 2

Step-by-step solution

∫ e sec ⁡ x ( sec ⁡ x tan ⁡ x f x + sec ⁡ x tan ⁡ x + sec 2 ⁡ x d x = e sec ⁡ x f x + C Differentiating both sides w.r.t ‘ x ’ we get e s e c x s e c x tan x f x + s e c x tan x + sec 2 x = e sec x ⋅ sec x tan x f ( x ) + e sec x f ' x ⇒ f ' x = s e c 2 x + t a n x s e c x ⇒ f x = ∫ sec 2 x + tan x sec x d x ⇒ f x = t a n x + s e c x + c ,   c ∈ R Hence, possible choice is f x = s e c x + t a n x + 1 2 .

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