Arc length x=2sint and y=2cost for 0 ≤ t ≤ 2π

Find the length of the curve given parametrically as x=2sint and y=2cost for 0 ≤ t ≤ 2π is the question .. any help would be awesome.

integrate this:

sqrt((dx/dt)^2 + (dy/dt)^2) * dt

From t = 0 to t = 2pi


x = 2 * sin(t)
dx/dt = 2 * cos(t)
(dx/dt)^2 = 4 * cos(t)^2

y = 2cos(t)
dy/dt = -2sin(t)
(dy/dt)^2 = 4 * sin(t)^2


(dx/dt)^2 + (dy/dt)^2 =>
4 * cos(t)^2 + 4 * sin(t)^2 =>
4 * (cos(t)^2 + sin(t)^2) =>
4 * 1 =>
4

sqrt(4) * dt =>
2 * dt

Integrate:

2t + C

From 0 to 2pi

2 * 2pi - 2 * 0 =>
4pi

x^2+y^2 = 4sin^2(t) + 4cos^2(t) = 4[sin^2(t)+cos^2(t)] = 4
Equation is a circle centre (0,0) radius 2
Total length of arc for 0 ≤ t ≤ 2π = 2π*2 = 4π

Arclength = ∫√(dx/dt)² + (dy/dt)² dt from a to b

In this problem:

Arclength = ∫2 dt from 0 to 2π = 4π