cos^2 α -sin^2 α =cos2 α [m] = ∫ ^{\frac{π}{3}}_{\frac{π}{6}}cos(2x+\frac{2π}{3})dx=\frac{1}{2}∫ ^{\frac{π}{3}}_{\frac{π}{6}}cos(2x+\frac{2π}{3})d(2x+\frac{2π}{3})=\frac{1}{2}\cdot sin((2x+\frac{2π}{3))|^{\frac{π}{3}}_{\frac{π}{6}}=...[/m] считайте