∫xcosx .dx=I
u=x, v´=cosx
u´=1, v=sinx
I=xsinx-∫sinxdx=xsinx-(-cosx)=xsinx+cosx
(∫uv´=uv-∫u´v)
Sin4x - cos⁴x = -sin⁴x
sin4x = cos⁴x - sin⁴x
sin4x = (cos²x - sin²x)(cos²x + sin²x)
sin4x = cos²x - sin²x
sin4x = cos2x
2sin2xcos2x - cos2x = 0
cos2x(2sin2x - 1) = 0
1) cos2x = 0
2x = π/2 + πn, n ∈ Z
x = π/4 + πn/2, n ∈ Z
2) 2sin2x - 1 = 0
sin2x = 1/2
2x = (-1)ⁿπ/6 + πk, k ∈ Z
x = (-1)ⁿπ/12 + πk/2, k ∈ Z
Ответ: x = π/4 + πn/2, n ∈ Z; (-1)ⁿπ/12 + πk/2, k ∈ Z.