Step 1: Recall the reduction formula.
For $I_n=\displaystyle\int\sin^n x\,dx$, the standard reduction formula is
\[ I_n=-\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}I_{n-2}. \]
Step 2: Move the $I_{n-2}$ term.
Rearranging,
\[ I_n-\frac{n-1}{n}I_{n-2}=-\frac{\sin^{n-1}x\cos x}{n}. \]
Step 3: Notice the matching shape.
The question asks for $I_6-\dfrac56 I_4$, which is exactly the left side with $n=6$ (since $\dfrac{n-1}{n}=\dfrac56$).
Step 4: Put $n=6$.
\[ I_6-\frac{5}{6}I_4=-\frac{\sin^{5}x\cos x}{6}. \]
Step 5: Check the powers.
$n-1=5$, so the power of $\sin$ is $5$ and there is one $\cos x$, with $6$ in the bottom.
Step 6: State the answer.
\[ \boxed{\dfrac{-\sin^{5}x\cos x}{6}} \]