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
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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