=(1/2)cos4x+(1/2)cos2x
sin2x*cosx*cos3x= (1/2)sin2x*cos4x+(1/2)sin2x*cos2x
формула sinα *cosβ=(1/2)sin(α+β)+(1/2)sin(α - β)
[b]sin2x*cosx*cos3x[/b]= (1/2)*sin2x*cos4x+(1/2)sin2x*cos2x=
(1/2)*(1/2)*sin(2x+4x)+ (1/2)*(1/2)*sin(2x-4x)+(1/4)2*sin2x*cos2x=
=(1/4)*sin6x-(1/4)*sin2x+(1/4)*sin4x
Интеграл от суммы равен сумме интегралов:
∫ sin2x*cosx*cos3x dx= ∫ (1/4)*sin6xdx- ∫(1/4)*sin2xdx + ∫(1/4)*sin4xdx=
= (1/24)*cos6x - (1/8)*cos2x +(1/16)*(-cos4x) + C=
= [b] (1/24)*cos6x - (1/8)*cos2x -(1/16)*cos4x + C[/b] - о т в е т.