If f(x) = ∫₀ˣ t(sin x - sin t) dt then?

f(x) = ∫₀ˣ t(sin x - sin t) dt = sin x ∫₀ˣ t dt - ∫₀ˣ t sin t dt = x²/2 sin x + t cos t |₀ˣ + sin x
f(x) = x²/2 sin x + x cos x + sin x
f'(x) = x²/2 cos x + 2 cos x
f''(x) = x cos x - x²/2 sin x - 2 sin x
f'''(x) = -x sin x + cos x - x sin x - x²/2 cos x - 2 cos x
Add f'''(x) and f'(x), we get
f'''(x) + f'(x) = cos x ± x sin x