Evaluate: ∫cos⁻¹(sinx)dx

Let cos⁻¹(sinx) = θ ⇒ sinx = cosθ ⇒ sinx = sin(π/2 - θ) ⇒ x = π/2 - θ ⇒ θ = π/2 - x
∴ ∫cos⁻¹(sinx)dx = ∫(π/2 - x)dx = ∫π/2dx - ∫xdx = πx/2 - x²/2 + c, where c is a constant of integration.