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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/2019/paper-1/20/c/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 / 2019 / Paper 1 / 20 / c / Solution by

Codex  0 2026-10-18

For the first definite integral, use integration by substitution

t = tan x

, with

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

,

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

and

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

. This gives

I = \int_{0}^{\infty} \frac{d t}{p + q t ^{2}} = \frac{1}{pq} [arctan (t \frac{q}{p})]_{0}^{\infty} = \frac{π}{2 pq} .

For the second calculation, let

D = p cos^{2} x + q sin^{2} x

. Since

p, q > 0

, this denominator is bounded away from zero; on compact positive parameter ranges the differentiated integrands have uniform bounds. Thus differentiation under the integral sign is justified and gives

I_{p} = - \int_{0}^{π /2} \frac{c o s ^{2} x}{D ^{2}} d x = - \frac{π}{4 p ^{3/2} q ^{1/2}}, I_{q} = - \int_{0}^{π /2} \frac{s i n ^{2} x}{D ^{2}} d x = - \frac{π}{4 p ^{1/2} q ^{3/2}} .

Adding the two equations and using the Pythagorean trigonometric identity in the numerators yields

J = - I_{p} - I_{q}

, hence

J = \frac{π ( p + q )}{4 ( pq ) ^{3/2}} .

For

p = q

, this becomes

π / (2 p^{2})

, agreeing with direct integration of the constant integrand.

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