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Let Ι 1 = ∫ 1 π 2 d t 1 + t 6 and Ι 2 = ∫ 0 π 2 x cos ⁡ x d x 1 + x s i n x + cos ⁡ x 6 , then

Options

  1. A2 Ι 1 = Ι 2
  2. BΙ 1 = 2 Ι 2
  3. CΙ 1 = Ι 2
  4. DΙ 1 = Ι 2 = 0

Correct answer

C. Ι 1 = Ι 2

Step-by-step solution

In Ι 2, let, x s i n x + cos ⁡ x = t ⇒ x c o s x + sin ⁡ x - sin ⁡ x d x = d t or x c o s x d x = d t ∴ Ι 2 = ∫ 1 π 2 d t 1 + t 6 = Ι 1 Also, Ι 1 & Ι 2 are both positive as 1 1 + t 6 > 0 ∀ t ∈ 1 , π 2

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