Step 1: Identify the type.
We need $\int_0^{\pi/2}\sin^6x\cos^4x\,dx$. Integrals of powers of sine and cosine from $0$ to $\tfrac{\pi}{2}$ are made for Wallis' formula.
Step 2: Recall the formula.
For even powers $m$ and $n$, \[ \int_0^{\pi/2}\sin^mx\cos^nx\,dx=\frac{(m-1)(m-3)\cdots1\cdot(n-1)(n-3)\cdots1}{(m+n)(m+n-2)\cdots2}\cdot\frac{\pi}{2}. \]
Step 3: List the factors.
Here $m=6$ and $n=4$. The sine part gives $5\cdot3\cdot1$ and the cosine part gives $3\cdot1$.
Step 4: Build the denominator.
Since $m+n=10$, the bottom is $10\cdot8\cdot6\cdot4\cdot2$.
Step 5: Put it together.
\[ \int_0^{\pi/2}\sin^6x\cos^4x\,dx=\frac{(5\cdot3\cdot1)(3\cdot1)}{10\cdot8\cdot6\cdot4\cdot2}\cdot\frac{\pi}{2}. \] The top is $15\times3=45$ and the bottom product is $3840$.
Step 6: Simplify.
$\dfrac{45}{3840}=\dfrac{3}{256}$, and multiplying by $\tfrac{\pi}{2}$ gives $\dfrac{3\pi}{512}$.
\[ \boxed{\dfrac{3\pi}{512}} \]