Ta có:
$\int\sin3x\cos5x\cos6xdx$
$=\int\dfrac{\sin3x(\cos11x+\cos x)}{2}dx$
$=\int\dfrac{\sin3x\cos11x+\sin3x\cos x}{2}dx$
$=\int\dfrac{\sin14x-\sin8x+\sin4x+\sin2x}{4}dx$
$=-\dfrac{\cos14x}{56}+\dfrac{\cos8x}{32}-\dfrac{\cos4x}{16}-\dfrac{\cos2x}{8}+C$