IMC 1998 · Day 1 · P6

Observe that the integral is equal to $$\int_0^{\frac{\pi}{2}} f(\sin\theta) \cos\theta\,d\theta$$ and to $$\int_0^{\frac{\pi}{2}} f(\cos\theta) \sin\theta\,d\theta$$ So, twice the integral is at most $$\int_0^{\frac{\pi}{2}} 1\,d\theta = \frac{\pi}{2}.$$ Now let $f(x) = \sqrt{1 - x^2}$. If $x = \sin\theta$ and $y = \sin\varphi$ then $$x f(y) + y f(x) = \sin\theta \cos\varphi + \sin\varphi \cos\theta = \sin(\theta + \varphi) \le 1.$$