换元积分法这是我的做法∵d( sin(x/2) ) = cos(x/2)/2 dx∴cos(x/2)dx = 2d( sin(x/2) )∴原式=∫ 2cosx d( sin(x/2) ) =2∫ 1-2sin^2(x/2) d( (sin(x/2) ) =2sin(x/2) - 2/3sin^3(x/2) + C 哪里错了...不好意思 打错了
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/19 14:11:08
![换元积分法这是我的做法∵d( sin(x/2) ) = cos(x/2)/2 dx∴cos(x/2)dx = 2d( sin(x/2) )∴原式=∫ 2cosx d( sin(x/2) ) =2∫ 1-2sin^2(x/2) d( (sin(x/2) ) =2sin(x/2) - 2/3sin^3(x/2) + C 哪里错了...不好意思 打错了](/uploads/image/z/3561673-49-3.jpg?t=%E6%8D%A2%E5%85%83%E7%A7%AF%E5%88%86%E6%B3%95%E8%BF%99%E6%98%AF%E6%88%91%E7%9A%84%E5%81%9A%E6%B3%95%E2%88%B5d%28+sin%28x%2F2%29+%29+%3D+cos%28x%2F2%29%2F2+dx%E2%88%B4cos%28x%2F2%29dx+%3D+2d%28+sin%28x%2F2%29+%29%E2%88%B4%E5%8E%9F%E5%BC%8F%3D%E2%88%AB+2cosx+d%28+sin%28x%2F2%29+%29++++++++++%3D2%E2%88%AB+1-2sin%5E2%28x%2F2%29+d%28+%28sin%28x%2F2%29+%29++++++++++%3D2sin%28x%2F2%29+-+2%2F3sin%5E3%28x%2F2%29+%2B+C+%E5%93%AA%E9%87%8C%E9%94%99%E4%BA%86...%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D+%E6%89%93%E9%94%99%E4%BA%86)
换元积分法这是我的做法∵d( sin(x/2) ) = cos(x/2)/2 dx∴cos(x/2)dx = 2d( sin(x/2) )∴原式=∫ 2cosx d( sin(x/2) ) =2∫ 1-2sin^2(x/2) d( (sin(x/2) ) =2sin(x/2) - 2/3sin^3(x/2) + C 哪里错了...不好意思 打错了
换元积分法
这是我的做法
∵d( sin(x/2) ) = cos(x/2)/2 dx
∴cos(x/2)dx = 2d( sin(x/2) )
∴原式=∫ 2cosx d( sin(x/2) )
=2∫ 1-2sin^2(x/2) d( (sin(x/2) )
=2sin(x/2) - 2/3sin^3(x/2) + C
哪里错了...
不好意思 打错了
更正下
原式=∫ 2cosx d( sin(x/2) )
=2∫ 1-2sin^2(x/2) d( (sin(x/2) )
=2sin(x/2) - 4/3sin^3(x/2) + C
换元积分法这是我的做法∵d( sin(x/2) ) = cos(x/2)/2 dx∴cos(x/2)dx = 2d( sin(x/2) )∴原式=∫ 2cosx d( sin(x/2) ) =2∫ 1-2sin^2(x/2) d( (sin(x/2) ) =2sin(x/2) - 2/3sin^3(x/2) + C 哪里错了...不好意思 打错了
1/3*sin(3/2*x)+sin(x/2)=1/3*(3sin(x/2)-4sin^3(s/2))+sin(x/2)=2sin(x/2)-4/3*sin(x/2)
所以答案其实是一样的