Area = integral y dx becomes

a^2 integral (1 - cos θ)^2 dθ

Area under one arch of a cycloid

Short answer

Integrate y dx from θ=0 to 2π: a²∫₀²π (1−cosθ)² dθ = 16a²∫₀^{π/2} sin⁴u du = 3πa² square units.

Find the area included between an arc of the cycloid x = a(θ − sinθ), y = a(1 − cosθ) and its base.

Cycloid: x = a(θ − sin θ), y = a(1 − cos θ). One arch is traced as θ goes from 0 to 2π; its base is the x-axis from x = 0 to x = 2πa.

dx = a(1 − cos θ) dθ.

Area = ∫ y dx
     = ∫₀^(2π) a(1 − cosθ) · a(1 − cosθ) dθ
     = a² ∫₀^(2π) (1 − cosθ)² dθ
     = a² ∫₀^(2π) 4 sin⁴(θ/2) dθ

Put θ/2 = u, dθ = 2du, limits 0 to π:

Area = 8a² ∫₀^π sin⁴u du = 16a² ∫₀^(π/2) sin⁴u du = 16a² · 3π/16

Area = 3πa² square units (three times the area of the generating circle).

✓ Answer checked by Claude Opus 5.5 (high effort)