第3个回答 2021-03-07
∫xsin²(8x)dx=(1/2)∫x[1-cos(16x)]dx=(1/2)[∫xdx-∫xcos(16x)dx]
=(1/2)[(1/2)x²-(1/16)∫xd[sin(16x)]=(1/4)x²-(1/32)[xsin(16x)-∫sin(16x)dx]
=(1/4)x²-(1/32)[xsin(16x)-(1/16)∫sin(16x)d(16x)]
=(1/4)x²-(1/32)[xsin(16x)+(1/16)cos(16x)]+C
=(1/4)x²-(1/32)xsin(16x)-(1/512)cos(16x)+C;