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AP EAMCET201825 Apr 2018Morning ShiftMathematicsIndefinite IntegrationActual

∫ cos 3 x + cos 5 x sin 2 x + sin 4 x d x =

Options

  1. Asin x - 6 tan - 1 ( sin x ) + c
  2. Bsin x - 2 ( sin x ) - 1 + c
  3. Csin x - 2 ( sin x ) - 1 - 6 tan - 1 ( sin x ) + c
  4. Dsin x - 2 ( sin x ) - 1 + 5 tan - 1 ( sin x ) + c

Correct answer

C. sin x - 2 ( sin x ) - 1 - 6 tan - 1 ( sin x ) + c

Step-by-step solution

Let, I = ∫ cos 3 x + cos 5 x sin 2 x + sin 4 x d x = ∫ cos 3 x 1 + cos 2 x sin 2 x 1 + sin 2 x d x = ∫ cos 2 x 1 + cos 2 x sin 2 x 1 + sin 2 x cos x d x = ∫ 1 - sin 2 x 1 + 1 - sin 2 x sin 2 x 1 + sin 2 x cos x d x = ∫ 1 - sin 2 x 2 - sin 2 x sin 2 x 1 + sin 2 x cos x d x Let sin x = t ⇒ cos x d x = d t I = ∫ 1 - t 2 2 - t 2 t 2 1 + t 2 d t = ∫ t 4 - 3 t 2 + 2 t 2 1 + t 2 d t = ∫ t 4 + t 2 - 4 t 2 + 2 t 2 1 + t 2 d t = ∫ t 4 + t 2 t 2 1 + t 2 d t + ∫

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