Calc. 2 question who can solve this

Calculate integral of 2sin^4 x cos^2xdx

∫2*sin^4(x)*cos^2(x) dx

2*∫(1 - cos(2x))²/4 * (1 + cos(2x))/2 dx

1/4*∫(1 - 2*cos(2x) + cos²(2x))*(1 + cos(2x)) dx

1/4*∫(1 - cos(2x) - cos²(2x) + cos^3(2x)) dx

x/4 - 1/8*sin(2x) - 1/8*∫(1 + cos(4x)) dx + 1/4*∫cos(2x)*(1 - sin²(2x)) dx

x/4 - 1/8*sin(2x) - x/8 - 1/32*sin(4x) + 1/8*[sin(2x) - sin^3(2x)/3] + C

x/8 - 1/8*sin(2x) - 1/32*sin(4x) + 1/8*sin(2x) - 1/24*sin^3(2x) + C

= x/8 - 1/32*sin(4x) - 1/24*sin^3(2x) + C

x = t/2
dx = dt/2

2 * sin(t/2)^4 * cos(t/2)^2 * (dt/2) =>
(sqrt((1 - cos(t))/2))^4 * (sqrt((1 + cos(t)) / 2))^2 * dt =>
(1/2)^2 * (1 - cos(t))^2 * (1/2) * (1 + cos(t)) * dt =>
(1/8) * (1 - cos(t)) * (1 + cos(t)) * (1 - cos(t)) * dt =>
(1/8) * (1 - cos(t)^2) * (1 - cos(t)) * dt =>
(1/8) * sin(t)^2 * (1 - cos(t)) * dt =>
(1/8) * sin(t)^2 - (1/8) * sin(t)^2 * cos(t) * dt

Now, let's integrate sin(t)^2

t = m/2
dt = dm/2

(1/8) * sin(m/2)^2 * (dm/2) =>
(1/16) * (1/2) * (1 - cos(m)) * dm =>
(1/32) * (1 - cos(m)) * dm

Integrate:

(1/32) * m - (1/32) * sin(m) =>
(1/32) * 2t - (1/32) * sin(2t) =>
(1/32) * 2 * (2x) - (1/32) * sin(2 * 2x) =>
(1/8) * x - (1/32) * sin(4x)

Let's integrate (1/8) * sin(t)^2 * cos(t) * dt

u = sin(t)
du = cos(t) * dt

(1/8) * u^2 * du

Integrate:
(1/24) * u^3 =>
(1/24) * sin(t)^3 =>
(1/24) * sin(x)^3

Now we have:

(1/8) * x - (1/32) * sin(4x) - (1/24) * sin(x)^3 + C