cosx^6的不定积分

Question

cosx^6的不定积分

Answer 1

主要是在化简指数问题上,尽量把次方形式变为复角形式,例如cos(nx)和sin(nx)等,比较好积
cos^6x
= (cos²)³
= [(1+cos2x)/2]³ = (1/8)(1+cos2x)³
= (1/8)(1+3cos2x+3cos²2x+cos³2x)
= (1/8)+(3/8)cos2x+(3/8)(1/2)(1+cos4x)+(1/8)(1/2)(1+cos4x)(cos2x)
= (1/8)+(3/8)cos2x+(3/16)+(3/16)cos4x+(1/16)(cos2x+cos4xcos2x)
= (5/16)+(3/8)cos2x+(3/16)cos4x+(1/16)cos2x+(1/32)cos6x+(1/32)cos2x
= (5/16)+(15/32)cos2x+(3/16)cos4x+(1/32)cos6x
∫cos^6x dx = ∫[(5/16)+(15/32)cos2x+(3/16)cos4x+(1/32)cos6x] dx
= (5/16)∫ dx + (15/32)∫cos2x dx + (3/16)∫cos4x dx + (1/32)∫cos6x dx
= (5/16)∫ dx + (15/32)(1/2)∫cos2x d(2x) + (3/16)(1/4)∫cos4x d(4x) + (1/32)(1/6)∫cos6x d(6x)
= (5/16)x + (15/64)sin2x + (3/64)sin4x + (1/192)sin6x + C