y=∫[sin^6x+x(2-cosx)/(x^4+cosx+10)]dx(其中积分上限为“兀/2”积分下限为“-兀/2”),求y

∫(-π/2,π/2) [sin^6x + x(2-cosx)/(x^4+cosx+10)]dx

=∫(-π/2,π/2) sin^6xdx + ∫(-π/2,π/2) x(2-cosx)/(x^4+cosx+10)dx

=∫(-π/2,π/2) sin^6xdx + 0

=2∫(0,π/2) sin^6xdx

=2·5/6·3/4·1/2·π/2

=5π/16

解题说明：∫(-π/2,π/2)表示以-π/2为下限，π/2为上限的定积分；

积分区间是一个对称区间，x(2-cosx)/(x^4+cosx+10)是一个奇函数，sin^6x是一个偶函数，分别可用对称区间上奇偶函数的积分性质。