Question

cos6x的原函数

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
=(cos_)_
=[(1+cos2x)/2]_=(1/8)(1+cos2x)_
=(1/8)(1+3cos2x+3cos_2x+cos_2x)
=(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^6xdx=∫[(5/16)+(15/32)cos2x+(3/16)cos4x+(1/32)cos6x]dx
=(5/16)∫dx+(15/32)∫cos2xdx+(3/16)∫cos4xdx+(1/32)∫cos6xdx
=(5/16)∫dx+(15/32)(1/2)∫cos2xd(2x)+(3/16)(1/4)∫cos4xd(4x)+(1/32)(1/6)∫cos6xd(6x)
=(5/16)x+(15/64)sin2x+(3/64)sin4x+(1/192)sin6x+C

Answer 2

∫cos6x dx
=(1/6)sin6x + C
原函数： (1/6)sin6x + C