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( s

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