Evaluate: ∫20√4−x2dx

Let I = ∫20√4−x2dx
I = ∫20√4−x2dx = ∫20√(2)2−x2dx = [x2√4−x2 + 42sin⁻¹x2]20 = (22×0 + 2sin⁻¹(1)) − (0 + 0) = 2sin⁻¹(1) = π
Hence ∫20√4−x2dx = π.