Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 11

\frac{3 x ^{2} + 5 x + 1}{x ( x + 1 ) ( x + 2 )} = \frac{1}{2 x} + \frac{1}{x + 1} + \frac{3}{2 ( x + 2 )} .

Integrating between one and two gives

\frac{1}{2} ln 2 + ln \frac{3}{2} + \frac{3}{2} ln \frac{4}{3} = \frac{1}{2} ln \frac{32}{3} .

All logarithm arguments are positive on this integration interval.

b

The double-angle identity gives

1 + cos 2 θ = 2 cos^{2} θ

. Therefore

\int_{0}^{π /4} \frac{d θ}{1 + cos 2 θ} = \frac{1}{2} \int_{0}^{π /4} sec^{2} θ d θ = \frac{1}{2} [tan θ]_{0}^{π /4} = \frac{1}{2} .

θ = π /4 - φ /2

, so

sin φ = cos 2 θ

and

d φ = - 2 d θ

. Reversing the transformed limits gives

\int_{0}^{π /2} \frac{d φ}{1 + sin φ} = 2 \int_{0}^{π /4} \frac{d θ}{1 + cos 2 θ} = 1.

The original integrand is finite at both endpoints.

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