Помогите пожайлуста срочно надо решить

1) sin (x)*sin (3x) так какsin (3x)=sin (2x+x)=sin (2x) cos (x)+sin (x) cos (2x), тоsin (x)*sin (3x)=sin (x)*[ sin (2x) cos (x)+sin (x) cos (2x) ]=sin (x)*[2sin (x) cos (x)*cos (x)+sin (x)*(2cos^2 (x) -1) ]=sin^2 (x)*[2cos^2 (x)+2cos^2 (x) -1]=sin^2 (x)*[4cos^2 (x) -1]=4sin^2 (x) cos^2 (x) -sin^2 (x) a. int (4sin^2 (x) cos^2 (x) dx=int (2sin (x) cos (x) ^2dx=int (sin (2x) ^2dx=int (1/2)*(1-cos (2*2x) dx=(1/2)*(x- (1/4)*sin (4x)+c б. int (sin^2 (x) dx=(-1/2) int (1-cos (2x) dx=(-1/2)*[x- (1/2) sin (2x) ]+c итогоint sin (x)*sin (3x) dx=(1/2)*[x- (1/4)*sin (4x) ]+c1+(-1/2)*[x- (1/2) sin (2x) ]+c2=(1/2)*[ (1/2) sin (2x) — (1/4) sin (4x) ]+c