设x^(1/2)=t
则x=t^2,dx=2tdt
代换得2∫(t^2sint)dt
(分部积分法)
2∫(t^2sint)dt
=2[-t^2cost+∫(2tcost)dt]
=2[-t^2cost+2tsint-∫sintdt]
=2[-t^2cost+2tsint+cost]
再代换得
(x^1/2)sin(x^1/2)dx
=-2xcos√x+4√xsin√x+2cos√x.