 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/13/b/solution

The argument

3 x^{3} + 2 x

is odd, so its squared sine is even. Replacing

x

- x

interchanges the numerator and denominator inside the logarithm, changing that factor's sign. Thus the full integrand is an odd function. It is continuous on the symmetric interval because

∣ x ∣ \leq 1/ π < 1

. Therefore

\int_{- 1/ π}^{1/ π} sin^{2} (3 x^{3} + 2 x) lo g \frac{1 - x ^{5}}{1 + x ^{5}} d x = 0 .

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