6∫0π|(sin3x+sin2x+sinx)|dx is equal to_____. [2026]
(17)
6∫0π|2sin2xcosx+sin2x|dx
=6∫0π|4sinxcos2x+2sinxcosx|dx
I=12∫0π|sinx(2cos2x+cosx)|dx
Put cosx=t, -sinx dx=dt
I=-12∫1-1|2t2+t|dt
I=12(∫-1-12(2t2+t)dt+∫-120-(2t2+t)dt+∫01(2t2+t)dt)
I=17