Question 73

Let $$I(x) = \int \frac{6}{\sin^2 x (1 - \cot x)^2} dx$$. If $$I(0) = 3$$, then $$I\left(\frac{\pi}{12}\right)$$ is equal to

Solution

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I(x) = ∫6/(sin²x(1-cotx)²)dx. Let u = 1-cotx, du = csc²x dx. I = ∫6/u² du = -6/u + C = -6/(1-cotx)+C. I(0): cotx→∞... I(π/12): cot(π/12) = 2+√3. 1-cot = -1-√3. I = -6/(-1-√3) = 6/(1+√3) = 6(√3-1)/2 = 3(√3-1). I(0)=3 condition: at x→0, -6/(1-cotx) → 0 (since cotx→∞), so C=3. I(π/12) = 3(√3-1)+3 = 3√3.

Option (3): 3√3.

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