Q.

 6∫0π|(sin3x+sin2x+sinx)|dx is equal to_____.    [2026]


Ans.

(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