Using sin3x+sinx=2sin2xcosx:
sin3x+sin2x+sinx=sin2x(2cosx+1)=2sinxcosx(2cosx+1)
Zeros in [0,π]: x=0,π/2,2π/3,π
Expression is positive on (0,π/2), negative on (π/2,2π/3), positive on (2π/3,π)
Let F(x)=−3cos3x−2cos2x−cosx
F(0)=−611, F(π/2)=21, F(2π/3)=125, F(π)=65
∫0π/2=37, ∫π/22π/3=121, ∫2π/3π=125
Total =37+121+125=617
6×617=17