★★★∫cos(x/2)cos(nx)dx (0→π)定积分★★★

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 19:00:45
★★★∫cos(x/2)cos(nx)dx (0→π)定积分★★★

★★★∫cos(x/2)cos(nx)dx (0→π)定积分★★★
★★★∫cos(x/2)cos(nx)dx (0→π)定积分★★★

★★★∫cos(x/2)cos(nx)dx (0→π)定积分★★★
利用积化和差公式
cos(x/2)cos(nx)=(1/2)[cos(n+1/2)x+cos(n-1/2)x]
积分=(1/2)∫[cos(n+1/2)x+cos(n-1/2)x]dx
=(1/(2n+1))sin[(n+1/2)x]+(1/(2n-1))sin[(n-1/2)x] |
=(1/(2n+1))sin[(n+1/2)π]+(1/(2n-1))sin[(n-1/2)π]
=(1/(2n+1))cosnπ-(1/(2n-1))cosnπ
=[(1/(2n+1))-(1/(2n-1))]cosnπ
=2*(-1)^(n+1)/(4n^2-1)