2011-2-13 22:29 满意回答 解：∵∫x?cos(4x)dx=x?sin(4x)/4-3/4∫ x?sin(4x)dx (应用分部积分法) =x?sin(4x)/4+3x?cos(4x)/16-3/8∫xcos(4x)dx (同上) =x?sin(4x)/4+3x?cos(4x)/16-3xsin(4x)/32+3/32∫ sin(4x)dx (同上) =x?sin(4x)/4+3x?cos(4x)/16-3xsin(4x)/32-3cos(4x)/128+C1 同理可得∫x?cos(2x)dx=x?sin(2x)/2+3x?cos(2x)/4-3xsin(2x)/4-3cos(2x)/8+C2 ∴∫ x?sin^4(x)dx=∫x?[3/8-cos(2x)/2+cos(4x)/8]dx (应用半角公式) =3/8∫x?dx-1/2∫x?cos(2x)dx+1/8∫x?cos(4x)dx =3x^4/32-1/2[x?sin(2x)/2+3x?cos(2x)/4-3xsin(2x)/4-3cos(2x)/8]+1/8[x?sin(4x)/4+3x?cos(4x)/16-3xsin(4x)/32-3cos(4x)/128]+C (C是积分常数) =3x^4/32+[sin(4x)/32-sin(2x)/4]x?+3[cos(4x)/128-cos(2x)/8]x?+3[sin(2x)/8-sin(4x)/256]x+3[cos(2x)/16-cos(4x)/1024]+C。