 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2/1/c/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

u = tan x

, one has

d x = d u / (1 + u^{2})

sin x = u / 1 + u^{2}

and

cos x = 1/ 1 + u^{2}

. The transformed improper integral is integrable at its zero endpoint and becomes

\int_{0}^{1/ 3} (u^{- 1/2} + u^{3/2}) d u = [2 u^{1/2} + \frac{2}{5} u^{5/2}]_{0}^{1/ 3} .

Hence

I = \frac{32}{15 4 3} .

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