∫([sin(6x-1)]/3)dx ? Just checking my work

[From: ] [author: ] [Date: 12-01-23] [Hit: ]

[-cos(6x-1)]/18+C

Could it be wrong? Thanks.

∫⅓sin(6x-1) dx=
=(1/18)∫sin(6x-1) d(6x-1)=
=-(1/18)cos(6x-1) +C

thats correct

keywords: Just,sin,checking,my,dx,int,work,∫([sin(6x-1)]/3)dx ? Just checking my work