∫u(x)dv(x)
=u(x) v(x)-∫v(x)du(x)
∫xsin xdx
=-∫xdcosx
u(x)=x v(x)=-cosx
所以
∫xsin xdx
=-∫xdcosx
=-[-xcosx-∫cosxdx]
=-[-xcosx-sinx+c]
=xcosx+sinx+c
c不分正负,最后只需+c
分部积分法求不定积分∫xsin xdx
∫u(x)dv(x)
=u(x) v(x)-∫v(x)du(x)
∫xsin xdx
=-∫xdcosx
u(x)=x v(x)=-cosx
所以
∫xsin xdx
=-∫xdcosx
=-[-xcosx-∫cosxdx]
=-[-xcosx-sinx+c]
=xcosx+sinx+c
c不分正负,最后只需+c