Let I(x) = ∫ 6 / (sin^2 x (1 - cot x)^2) dx. If I(0) = 3, then I(π/12) is equal to :
Step-by-Step Solution
Key Concept: Use substitution u = 1 - cot x, then du = cosec^2 x dx = (1/sin^2 x) dx. The integral becomes \int 6 / u^2 du = -6/u + C = -6 / (1 - cot x) + C.
Let u = 1 - cot x, then du = cosec^2 x dx. The integral becomes \int 6 du / u^2 = -6/u + C = -6 / (1 - cot x) + C. Given I(0) = 3, we evaluate the limit as x approaches 0. As x \to 0, cot x \to \infty, so -6 / (1 - cot x) \to 0. Thus, C = 3. So I(x) = 3 - 6 / (1 - cot x). At x = \pi/12, cot(\pi/12) = cot(15°) = 2 + \sqrt{3.} I(\pi/12) = 3 - 6 / (1 - (2 + \sqrt{3})) = 3 - 6 / (-1 - \sqrt{3}) = 3 + 6 / (1 + \sqrt{3}) = 3 + 6(\sqrt{3} - 1) / 2 = 3 + 3(\sqrt{3} - 1) = 3 + 3\sqrt{3} - 3 = 3\sqrt{3.}
Correct Answer: (2)