∫(sinx)^4dx+∫(sinx)^6dx

记A＝∫(0到π) x(sinx)^6dx,换元x＝π-t,则A＝∫(0到π) π(sint)^6dt－∫(0到π) t(sint)^6dt,所以A＝π/2×∫(0到π) (sinx)^6dx.
(sinx)^6以π为周期,且是偶函数,所以∫(0到π) (sinx)^6dx＝∫(-π/2到π/2) (sinx)^6dx＝2∫(0到π/2) (sinx)^6dx,套用定积分公式,∫(0到π) (sinx)^6dx＝2×5/6×3/4×1/2×π/2
所以,原积分A＝π/2×2×5/6×3/4×1/2×π/2＝5π^2/32