Will someone help me evaluate ∫xsin^2(x)cos(x)dx

(1/3)xsin^3(X) + (1/3)cos(x) - (1/9)cos^3(x) + c

I only managed to get the (1/3)xsin^3(X) in my answer...

Integrate by parts:
u = x . . . . . . dv = sin²x cosx dx
du = dx . . . . . v = 1/3 sin³x

∫ u dv = u v − ∫ v du

∫ x sin²x cosx dx = (1/3) x sin³x − ∫ (1/3) sin³x dx
. . . . . . . . . . . . = (1/3) x sin³x − ∫ (1/3) sinx (1 − cos²x) dx
. . . . . . . . . . . . = (1/3) x sin³x − ∫ (1/3) sinx dx + ∫ (1/3) sinx cos²x dx
. . . . . . . . . . . . = (1/3) x sin³x + (1/3) cosx − (1/9) cos³x + C

Integrate by parts:

int(u * dv) = u * v - int(v * du)

u = x
du = dx
dv = sin(x)^2 * cos(x) * dx
v = (1/3) * sin(x)^3

(1/3) * x * sin(x)^3 - int((1/3) * sin(x)^3 * dx) =>
(1/3) * x * sin(x)^3 - (1/3) * int(sin(x)^2 * sin(x) * dx) =>
(1/3) * x * sin(x)^3 - (1/3) * int((1 - cos(x)^2) * sin(x) * dx) =>
(1/3) * x * sin(x)^3 - (1/3) * int(sin(x) * dx) + (1/3) * int(cos(x)^2 * sin(x) * dx)

u = cos(x)
du = -sin(x) * dx

(1/3) * x * sin(x)^3 + (1/3) * cos(x) + (1/3) * int(u^2 * -du) =>
(1/3) * x * sin(x)^3 + (1/3) * cos(x) - (1/3) * int(u^2 * du) =>
(1/3) * x * sin(x)^3 + (1/3) * cos(x) - (1/3) * (1/3) * u^3 + C =>
(1/3) * x * sin(x)^3 + (1/3) * cos(x) - (1/9) * cos(x)^3 + C