请教如何解答？∫xcos（x/2）dx

用分部积分
∫xcos（x/2）dx
=2∫xcos(x/2)d(x/2)
=2∫xdsin(x/2)
=2xsin(x/2)-2∫sin(x/2)dx
=2xsin(x/2)-4∫sin(x/2)d(x/2)
=2xsin(x/2)+4cos(x/2)+C

∫xcos（x/2）dx
=2∫xcos(x/2)d(x/2)
=2∫xdsin(x/2)
=2xsin(x/2)-2∫sin(x/2)dx
=2xsin(x/2)-4∫sin(x/2)d(x/2)
=2xsin(x/2)＋4cos(x/2)＋c
PS:∫dx=2∫d(x/2) ＋c
∫sinxdx=-cosx ＋c
∫cosxdx=sinx ＋c(c为任意的常数，c'=0)
分部积分法：设两个被积函数为U(x)和V(x)
∫U(x)dV(x)=U(x)V(x)-∫V(x)dU(x)