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AP EAMCET20224 Jul 2022Evening ShiftMathematicsIndefinite IntegrationActual

If f x = ∫ x 2 cos 2 x 2 x tan 2 x - 2 x - 6 tan x d x and f 0 = π then f x =

Options

  1. Ax 2 sin x + π
  2. Bcos x + π - 1
  3. C- x 3 sin 2 x + π
  4. Dx 3 cos 2 x + π cos x

Correct answer

C. - x 3 sin 2 x + π

Step-by-step solution

f x = ∫ x 2 cos 2 x 2 x tan 2 x - 2 x - 6 tan x d x = ∫ 2 x 3 sin 2 x - 2 x 3 cos 2 x - 6 x 2 sin x cos x d x = ∫ 2 x 3 - cos 2 x d x - ∫ 3 x 2 sin 2 x d x Let I = - ∫ 2 x 3 cos 2 x d x = - 2 x 3 sin 2 x 2 + 2 ∫ 3 x 2 sin 2 x 2 d x = - x 3 sin 2 x + ∫ 3 x 2 sin 2 x d x Hence, f x = - x 3 sin 2 x + C Given, f 0 = π i.e. C = π ⇒ f x = - x 3 sin 2 x + π

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