∫x?cosxdx=∫x?d(sinx)=x?sinx-∫sinxd(x?)=x?sinx-2∫xsinxdx=x?sinx+2∫xd(cosx)=x?sinx+2xcosx-2∫cosxdx=x?sinx+2xcosx-2sinx+C (C为任意常数)
求x平方cosxdx的不定积分的计算过程
∫x?cosxdx=∫x?d(sinx)=x?sinx-∫sinxd(x?)=x?sinx-2∫xsinxdx=x?sinx+2∫xd(cosx)=x?sinx+2xcosx-2∫cosxdx=x?sinx+2xcosx-2sinx+C (C为任意常数)