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NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice

The integral I = ∫ 2 sin ⁡ x 3 + sin ⁡ 2 x d x simplifies to (where, C is the constant of integration)

Options

  1. Aln ⁡ 2 + sin ⁡ x - cos ⁡ x 2 - sin ⁡ x + cos ⁡ x - tan - 1 ⁡ sin ⁡ x + cos ⁡ x + C
  2. Bln ⁡ ( sin ⁡ x ) + sin ⁡ 2 x + C
  3. Csin ⁡ 2 x - ln ⁡ ( cos ⁡ x ) + C
  4. D1 4 ln ⁡ 2 + sin ⁡ x - cos ⁡ x 2 - sin ⁡ x + cos ⁡ x - 1 2 tan - 1 ⁡ sin ⁡ x + cos ⁡ x 2 + C

Correct answer

D. 1 4 ln ⁡ 2 + sin ⁡ x - cos ⁡ x 2 - sin ⁡ x + cos ⁡ x - 1 2 tan - 1 ⁡ sin ⁡ x + cos ⁡ x 2 + C

Step-by-step solution

The given integral can be rewritten as, I = ∫ sin ⁡ x + cos ⁡ x 3 + sin ⁡ 2 x d x + ∫ sin ⁡ x - cos ⁡ x 3 + sin ⁡ 2 x d x Let sin ⁡ x - cos ⁡ x = t and sin ⁡ x + cos ⁡ x = u in 1 s t and 2 n d integral respectively. ∴ I = ∫ d t 4 - t 2 - ∫ d u 2 + u 2 = 1 4 ln ⁡ 2 + t 2 - t - 1 2 tan - 1 ⁡ u 2 + C Putting values of t and u , we get, I = 1 4 ln ⁡ 2 + sin x - cos x 2 - sin x + cos x - 1 2 tan - 1 ⁡ sin x

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