Question

求定积分∫（e,1）sinlnxdx

Answer 1

∫sinlnxdx
=xsinlnx-∫xcoslnx*(1/x)dx
=xsinlnx-∫coslnxdx
=xsinlnx-[xcoslnx-∫x*(-sinlnx)*(1/x)dx]
=xsinlnx-xcoslnx-∫sinlnxdx
移项,得：2∫sinlnxdx=xsinlnx-xcoslnx+C'
∫sinlnxdx=(1/2)(xsinlnx-xcoslnx)+C
∫(1→e)sinlnxdx
=[(1/2)(xsinlnx-xcoslnx)]|(1→e)
=(1/2)(esinlne-ecoslne)-(1/2)(1sinln1-1cosln1)
=(1/2)(esin1-ecos1)-(1/2)(1sin0-1cos0)
=(e/2)(sin1-cos1)+1/2