 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/7a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Use the suggested substitution

x = \frac{t}{2 - t ^{2}} .

[0, 1]

to itself and satisfies

d x = \frac{2 d t}{( 2 - t ^{2} ) ^{3/2}}, 1 - x^{4} = \frac{4 ( 1 - t ^{2} )}{( 2 - t ^{2} ) ^{2}} .

Hence

\int_{0}^{1} \frac{d x}{1 - x ^{4}} = \frac{1}{2} \int_{0}^{1} \frac{d t}{( 1 - t ^{2} ) ( 1 - t ^{2} /2 )} = \frac{1}{2} K (1/ 2),

where

K

Γ (5/4) = Γ (1/4) /4

now give

K (1/ 2) = \frac{Γ ( 1/4 ) ^{2}}{4 π} = \frac{4Γ ( 5/4 ) ^{2}}{π} .

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