 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-4/14e/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-18

t = x^{2}

, so

d x = d t / (2 t)

. Then

I = \frac{1}{2} \int_{0}^{1} t^{- 1/3} (1 - t)^{1/3} d t = \frac{1}{2} B (2/3, 4/3) .

I = \frac{1}{2} Γ (2/3) Γ (4/3) = \frac{1}{6} Γ (2/3) Γ (1/3) = \frac{π}{6 sin ( π /3 )} .

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