 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2/section-b/2/solution

t = tan (x /2)

gives

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

and

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

. The endpoints

x = 0, π /2

become

t = 0, 1

, respectively. Hence

\int_{0}^{π /2} \frac{d x}{2 + s i n x} = \int_{0}^{1} \frac{2 d t}{2 ( 1 + t ^{2} ) + 2 t} = \int_{0}^{1} \frac{d t}{t ^{2} + t + 1} .

t^{2} + t + 1 = (t + 1/2)^{2} + 3/4

, then gives

\int_{0}^{1} \frac{d t}{t ^{2} + t + 1} = \frac{2}{3} [arctan \frac{2 t + 1}{3}]_{0}^{1} = \frac{2}{3} (\frac{π}{3} - \frac{π}{6}) = \frac{π}{3 3} .

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