不定积分自己算完和答案使用的公式不同,得出不同结果,不知道是不是我算错了,

cosαcosβ=[cos（α+β）+cos（α-β）]/2
cosxcosx/2=[(cos(3x/2)+cosx/2)/2
原式=f (cos(3x/2)/2dx+f (cosx/2)/2dx
=sin(3x/2)/3+sin(x/2)+C
微分得:
(sin(3x/2)/3+sin(x/2)+C)'
cos(3x/2)/2+cos(x/2)/2
=(cos3x/2+cosx/2)/2
=cosxcosx/2
f cosxcosx/2dx
=f (2cos^2(x/2)-1)cosx/2dx
=2f (2(1-sin^2(x/2)-1)d(sinx/2)
=2f(1-2sin^2(x/2))d(sinx/2)
=2sinx/2-4/3sin^3(x/2)+C (这里,你做错了,不是8/3sin^3(x/2))
微分得:
(2sinx/2-4/3sin^3(x/2)+C)'
=cos(x/2)-3*4/3sin^2(x/2)cos(x/2) /2
=cos(x/2)-2sin^2(x/2)cos(x/2)
=cos(x/2)*(1-2sin^2(x/2)
=cos(x/2)cosx

4∫sin²(x/2)dsin(x/2)
这里是4*1/3*sin³(x/3)+C
不是2/3
∫x²dx=x³/3+C