If ##f(x) = \arcsin(x)/(1+x^8)##, the possible case ##\int_{-1}^1 f(x) \, dx## is easy to compute explicity (with almost no work), and the inequality is (trivially) true in that case. The other obvious possibility, ##\int_0^1 f(x) \, dx##, takes more work; the lower bound is easy to verify (again, with essentially no work), while the upper bound is true (as seen numerically), but is not at all obvious anymore. Why don't you see if you can prove ##\int_0^1 f(x) \, dx \leq \frac{\pi}{2} - 1##, without performing a numerical evaluation of the integral?
