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ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2012/6/a/solution

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Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 6 / a / Solution by

Codex  0 2026-10-31

The tangent half-angle substitution

u = tan (t /2)

0 \leq t < π

to

0 \leq u < \infty

, with

cos t = \frac{1 - u ^{2}}{1 + u ^{2}}, d t = \frac{2 d u}{1 + u ^{2}} .

Because

a > 1

, both

a + 1

and

a - 1

are positive. Thus

\int_{0}^{π} \frac{d t}{a + c o s t} = 2 \int_{0}^{\infty} \frac{d u}{( a + 1 ) + ( a - 1 ) u ^{2}} = \frac{2}{a ^{2} - 1} [arctan (u \frac{a - 1}{a + 1})]_{0}^{\infty} .

Evaluating the improper integral gives

\frac{π}{a ^{2} - 1} .

Although the change of variable has an infinite endpoint, the original integrand is continuous on the closed interval, and the transformed integrand decays as

u^{- 2}

, so the limiting substitution is justified.

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