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JEE Main202427 Jan 2024Evening ShiftMathematicsDefinite IntegrationActual

For 0 < a < 1 , the value of the integral ∫ 0 π d x 1 - 2 a cos x + a 2 is :

Options

  1. Aπ 2 π + a 2
  2. Bπ 2 π - a 2
  3. Cπ 1 - a 2
  4. Dπ 1 + a 2

Correct answer

C. π 1 - a 2

Step-by-step solution

Given: I = ∫ 0 π dx 1 - 2 acosx + a 2 . . . i ⇒ I = ∫ 0 π dx 1 - 2 acos π - x + a 2 ⇒ I = ∫ 0 π dx 1 + 2 acosx + a 2 . . . i i Adding i and i i , ⇒ 2 I = ∫ 0 π dx 1 - 2 acosx + a 2 + ∫ 0 π dx 1 + 2 acosx + a 2 ⇒ 2 I = ∫ 0 π 1 1 - 2 acosx + a 2 + 1 1 + 2 acosx + a 2 dx ⇒ 2 I = ∫ 0 π 1 + 2 acosx + a 2 + 1 - 2 acosx + a 2 1 + a 2 2 - 2 acosx 2 dx ⇒ 2 I = ∫ 0 π 2 1 + a 2 1 + a 2 2 - 2 acosx 2 dx ⇒ 2 I = 2 ∫ 0 π 2 2 1 + a 2 1 + a 2 2 - 4 a 2 cos 2 x dx ⇒ I = ∫ 0 π 2 2 1 + a 2 · sec 2 x 1 + a 2 2 · sec 2 x - 4 a 2 dx ⇒ I

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