Step 1: Rewrite the integrand
Since $\sin x\cos x=\frac{\sin2x}{2}$, the integrand is $\frac{1}{8}\sin^{3}2x$, and the candidate answer can be checked by differentiation.
Step 2: Check by differentiation
Differentiate $-\frac{1}{192}(9\cos2x-\cos6x)$: we get $-\frac{1}{192}(-18\sin2x+6\sin6x)=\frac{18\sin2x-6\sin6x}{192}=\frac{3\sin2x-\sin6x}{32}$.
Step 3: Match
Since $3\sin2x-\sin6x=4\sin^{3}2x$, the derivative is $\frac{\sin^{3}2x}{8}=\sin^{3}x\cos^{3}x$, as required.
Final Answer:
The first option differentiates back to the integrand.
\[ \boxed{\text{(A) }-\dfrac{1}{192}[9\cos 2x-\cos 6x]+c} \]