∫（上限π／2 下限0） [(sint)^4-(sint)^6] dt

当n为正整偶数时,即n=2m,m=1,2...
∫(0→π/2)(sinx)^ndx=[(2m-1)!/(2m)!](π/2)
当n为正整奇数时,即n=2m+1,m=0,1,2...
∫(0→π/2)(sinx)^ndx=[(2m)!/(2m+1)!]
∫(0→π/2)(sinx)^4dx
=(3/4)×(1/2)×(π/2)
=3π/16
∫(0→π/2)(sinx)^6dx
=(5/6)×(3/4)×(1/2)×(π/2)
=5π/32
3π/16-5π/32=π/32