∫(上限π/2 下限0) [(sint)^4-(sint)^6] dt
问题描述:
∫(上限π/2 下限0) [(sint)^4-(sint)^6] dt
答
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...