Step 1: Identify where the signs change.
Over $\left[\frac{\pi}{4},\frac{5\pi}{4}\right]$, $\cos t$ is positive on $\left[\frac{\pi}{4},\frac{\pi}{2}\right]$ and negative on $\left[\frac{\pi}{2},\frac{5\pi}{4}\right]$, while $\sin t$ is positive on $\left[\frac{\pi}{4},\pi\right]$ and negative on $\left[\pi,\frac{5\pi}{4}\right]$.
Step 2: Set up the integral on $\left[\frac{\pi}{4},\frac{\pi}{2}\right]$.
Here both are positive, so the integrand $|\cos t|\sin t+|\sin t|\cos t=\cos t\sin t+\sin t\cos t=2\sin t\cos t=\sin 2t$.
Step 3: Set up on $\left[\frac{\pi}{2},\pi\right]$.
Here $\cos t<0$, $\sin t>0$, so $|\cos t|\sin t=-\cos t\sin t$ and $|\sin t|\cos t=\sin t\cos t$; they cancel, integrand $=0$.
Step 4: Set up on $\left[\pi,\frac{5\pi}{4}\right]$.
Here both negative, so $|\cos t|\sin t=-\cos t\sin t$ and $|\sin t|\cos t=-\sin t\cos t$, summing to $-2\sin t\cos t=-\sin 2t$.
Step 5: Integrate the nonzero pieces.
First piece: $\int_{\pi/4}^{\pi/2}\sin 2t\,dt=\left[-\frac{\cos 2t}{2}\right]_{\pi/4}^{\pi/2}=-\frac{\cos\pi}{2}+\frac{\cos(\pi/2)}{2}=\frac{1}{2}$. Third piece: $\int_{\pi}^{5\pi/4}(-\sin 2t)\,dt=\left[\frac{\cos 2t}{2}\right]_{\pi}^{5\pi/4}=\frac{\cos(5\pi/2)}{2}-\frac{\cos 2\pi}{2}=0-\frac{1}{2}=-\frac{1}{2}$.
Step 6: Add all contributions.
Total $=\frac{1}{2}+0+\left(-\frac{1}{2}\right)=0$, matching the key. \[ \boxed{0} \]