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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/13/e/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 13 / e / Solution by

Codex  0 2026-10-18

Use the tangent half-angle substitution

t = tan (x /2)

, so

cos x = (1 - t^{2}) / (1 + t^{2})

and

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

. Then

\int \frac{d x}{5 + 3 c o s x} = \int \frac{2 d t}{5 ( 1 + t ^{2} ) + 3 ( 1 - t ^{2} )} = \int \frac{d t}{t ^{2} + 4} .

The inverse tangent antiderivative gives

\frac{1}{2} arctan (\frac{1}{2} tan \frac{x}{2}) + C .

This expression applies on each interval

((2 j - 1) π, (2 j + 1) π)

for an integer

j

. Although

5 + 3 cos x \geq 2

everywhere, the chosen inverse tangent expression jumps at the endpoints because the substitution has a pole there. A global continuous antiderivative is obtained by adding

jπ /2

on that interval, using one common

C

, and extending by continuity at the endpoints.

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