第1个回答 2012-11-10
cosxcos2xcos3x
= cos2x * cosxcos3x
= cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)]
= cos2x * (1/2)[cos4x + cos2x]
= (1/2)cos2xcos4x + (1/2)cos²2x
= (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x)
= (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4x
∫ cosxcos2xcos3x dx
= (1/4)∫ dx + (1/4)∫ cos2x dx + (1/4)∫ cos4x dx + (1/4)∫ cos6x dx
= x/4 + (1/8)sin2x + (1/16)sin4x + (1/24)sin6x + C本回答被网友采纳