企源知识库

求∫sin√xdx
【正确答案】：令√x=u，则x=u²(u≥0)，dx=2udu，于是 ∫sin√xdx=∫sinu•2udu =2∫nsinudu =-2∫ud(cosu) =-2(ucosu-∫cosudu) =-2(ucosu-sinu)+C =2(sin√x-√xcos√x)+C．