Найти первообразную для функции: f (x) = 2sin^2x f (x) = sin2x*cos2x

f (x) = 2sin2x-2

F (x) = 2*1/2 * (-cos2x) - 2x = - cos2x-2x

1) ∫ (2sin²x) dx=∫ (1-cos2x) dx=∫dx-∫cos2xdx=x-1/2*∫cos2x * d (2x) = x-1/2 * sin2x+C

2) ∫sin2x * cos2x dx=∫ 1/2 * sin4x dx=1/2 * 1/4 * ∫sin4x * d (4x) = 1/8 * (-cos4x) + C=-1/8cos4x+C