If g(x) = ∫sin(2x)sin(x)sin⁻¹(t)dt, then g'(π/2) = ? g'(-π/2) = ?

Remember sin⁻¹(sin x) = x ∀ x ∈ [-π, π]
g(x) = ∫sin(2x)sin(x)sin⁻¹(t)dt
∴g'(x) = sin⁻¹(sin(2x)) × cos(2x) × 2 - sin⁻¹(sin(x)) × cos(x)
⇒g'(π/2) = sin⁻¹(sin(2π/2)) × cos(2π/2) × 2 - sin⁻¹(sin(π/2)) × cos(π/2)
g'(π/2) = sin⁻¹(sin(π)) × cos(π) × 2 - sin⁻¹(sin(π/2)) × cos(π/2)
⇒g'(π/2) = π
Also,
g'(-π/2) = sin⁻¹(sin(2(-π/2))) × cos(2(-π/2)) × 2 - sin⁻¹(sin(-π/2)) × cos(-π/2)
g'(-π/2) = sin⁻¹(sin(-π)) × cos(-π) × 2 - sin⁻¹(sin(-π/2)) × cos(-π/2)
⇒g'(-π/2) = 2π