Set up, but do not evaluate, the integral that represents the length of the curve given by

x = t + cos (t), y = t -sin (t) over the interval 0 ≤ t ≤ 2π.

Thank you!

L = ∫₀²ᵖⁱ √((1 − sin t)² + (1 − cos t)²) dt

Step-by-step explanation:

Arc length of a parametric curve is:

L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt

x = t + cos t, dx/dt = 1 − sin t

y = t − sin t, dy/dt = 1 − cos t

L = ∫₀²ᵖⁱ √((1 − sin t)² + (1 − cos t)²) dt

Or, if you wish to simplify:

L = ∫₀²ᵖⁱ √(1 − 2 sin t + sin²t + 1 − 2 cos t + cos²t) dt

L = ∫₀²ᵖⁱ √(3 − 2 sin t − 2 cos t) dt

answer: options

step-by-step explanation: