∫(x^2)cos（x⼀2）dx用分部积分法

如果答案是(1/6)x^3+(1/2)(x^2)sinx+xcosx-sinx+C，那么你的题目抄错咯，题目应该是：
∫x²cos²(x/2) dx，那么
∫ x²cos²(x/2)dx
=∫1/2x²(1+cosx)dx
=1/2∫ x²dx+∫1/2x²cosxdx
=1/6 x³+1/2∫x²dsinx
=1/6 x³+1/2 x²sinx-∫xsinxdx
=1/6 x³+1/2 x²sinx-[-xcosx+∫cosxdx]
=1/6 x³+1/2 x²sinx+ xcosx-sinx+C
如果题目是∫ x²cos(x/2)dx，那么，参考答案错了，正确的答案是：
∫ x²cos(x/2)dx=∫2x²dsin(x/2)
= 2x²sin(x/2)- ∫2 sin(x/2) dx²
=2x²sin(x/2)- ∫4xsin(x/2) dx
=2x²sin(x/2)- ∫-8x dcos(x/2)
=2x²sin(x/2)+8xcos(x/2)- ∫8cos(x/2)dx
=2x²sin(x/2)+8xcos(x/2)- 16sin(x/2)+C

∫x²cos(x/2) dx
=2∫x²cos(x/2) d(x/2)=2∫x² d[sin(x/2)]
=2x²sin(x/2)-2∫sin(x/2) d(x²)
=2x²sin(x/2)-4∫xsin(x/2) dx
=2x²sin(x/2)-8∫x d[-cos(x/2)]
=2x²sin(x/2)+8xcos(x/2)+8∫cos(x/2) dx
=2x²sin(x/2)+8xcos(x/2)+16∫cos(x/2) d(x/2)
=2x²sin(x/2)+8xcos(x/2)+16sin(x/2)+C
=2(x²-8)sin(x/2)+8xcos(x/2)+C