 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/17/a/i/solution

u = cos x

d u = - sin x d x

. The denominator factorizes as

(u - 2) (u - 3)

- \int \frac{d u}{( u - 2 ) ( u - 3 )} = \int (\frac{1}{u - 2} - \frac{1}{u - 3}) d u .

Integrating and restoring

x

yields

\int \frac{sin x}{cos ^{2} x - 5 cos x + 6} d x = ln \frac{2 - cos x}{3 - cos x} + C .

For real

x

, both

2 - cos x

and

3 - cos x

are positive, so no absolute-value signs are needed in this real logarithm. The denominator never vanishes.

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