∫(0→π/2) [(sint)^4－(sint)^6]
问题描述：
∫(0→π/2) [(sint)^4－(sint)^6]
答
估计你的书本应该有这样一条公式：
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)!]
知道这些后就好办了
∫(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
所以,原式=3π/16-5π/32=π/32