∫(-π/4,π/4) (x^3+cosx)/[1+sin^(2)x]dx
=∫(-π/4,π/4) x^3/[1+sin^(2)x]dx+∫(-π/4,π/4) cosx/[1+sin^(2)x]dx（前者奇函数,后者偶函数）
=0+2∫(0,π/4) d(sinx)/[1+sin^(2)x]
=2∫(0,√2/2) d(t)/[1+t^2]
=2tan^(-1)t|(0,√2/2)
=2tan^(-1)(√2/2)