Solution

First of all, observe that

Therefore, summing up the two and using the well known identity, we get that their sum equals 2π. In particular,

This solves the problem. However, several integrals may be obtained with no additional effort. For one, the second integral above is also π. Now, since the period of sin²(x) is actually π, we easily get another integral

Finally, taking into account the symmetry of sin, sin(π - x) = sin(x), we also obtain

On a flippant note, every time we halved the interval of integration, the integral got halved as well. Would it be right to expect that the next time we'll get an integral equal to one eighth of π?

In a similar vein, consider the following integral:

The values taken by the integrand f(x), as x runs from 0 to π/2, are exactly the same as taken on by f(π/2 - x) on the same interval, except that they are generated in reverse order. Consequently, the integrals of the two functions are equal:

Summing up,