> 数学 >
∫(0→π/2) [(sint)^4-(sint)^6]
人气:159 ℃ 时间:2019-12-21 16:41:10
解答
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
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版