Verified answer

∫ x⁵ cos(x³) dx =====> substitution

u = x^3

du = 3x^2 dx ====> du / 3x^2 = dx

∫ x^3 * x^2 cos(u) du / 3x^2

(1/3) * ∫ x^3 * cos(u) du

(1/3) * ∫ u * cos(u) du ( by parts )

s = u ...........dv = cos(u) du

ds = du.........v = sin(u)

s * v - ∫ v * ds

(1/3) * [ u * sin(u) - ∫ sin(u) du ]

(1/3) * [ u * sin(u) + cos(u) ]

(1/3)*u * sin(u) + (1/3) * cos(u) + C

(1/3)* x^3 * sin(x^3) + (1/3) * cos(x^3) + C

so you have

u = x^3

∫x cos(x)dx = [x sin x] - ∫sin(x) dx = x sin x - cos x

is the route you are then going down, give or take factors of course

integration by parts

∫uv' = uv - ∫u'v