试证:∫(0->π/2)cos^n xdx=∫(0->π/2)sin^n xdx
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:综合作业 时间:2024/08/03 15:53:09
试证:∫(0->π/2)cos^n xdx=∫(0->π/2)sin^n xdx
从"于是"后面起第二行是cos^n(π/2)怎么变成的sin^nt
![](http://img.wesiedu.com/upload/2/35/235bbd75408cc8e842f7e7c6d7356f09.jpg)
从"于是"后面起第二行是cos^n(π/2)怎么变成的sin^nt
![](http://img.wesiedu.com/upload/2/35/235bbd75408cc8e842f7e7c6d7356f09.jpg)
![试证:∫(0->π/2)cos^n xdx=∫(0->π/2)sin^n xdx](/uploads/image/z/20016165-21-5.jpg?t=%E8%AF%95%E8%AF%81%3A%E2%88%AB%280-%3E%CF%80%2F2%29cos%5En+xdx%3D%E2%88%AB%280-%3E%CF%80%2F2%29sin%5En+xdx)
中学时学的公式忘记了吗?
sin(π/2-x)=cosx
cos(π/2-x)=sinx
sin(π/2-x)=cosx
cos(π/2-x)=sinx
证明∫(0,π/2)sin^m x cos^m x dx=1/2^m∫(0,π/2)cos^m xdx
求定积分∫上限π下限0 cos xdx
计算定积分:∫(0,π) cos²xdx
定积分【0,π^2】sin根号xdx
求定积分∫上限π/2,下限0 4sin^2xcos^2xdx,
计算不定积分∫x^21n xdx
求定积分!∫(-π,π)√(1+cos2x)+cosx^2sin^3xdx
不定积分∫根号下tanx+1/cos^2xdx
求不定积分∫e^(-x)cos^2xdx
问高数求导 ∫sin^3xcos^2xdx
求定积分∫(sinx)^(n-1)cos(n+1)xdx,上限为π,下限为0.书上说用分部积分法
∫根号xdx=,