To solve the given problem, we need to evaluate the following expression:
\[
\int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx + \int_{2\pi}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx
\]
### Step 1: Evaluate the first integral
\[
\int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx
\]
The integral can be split into two parts:
\[
\int_{0}^{\frac{\pi}{4}} \cos x \, dx - \int_{0}^{\frac{\pi}{4}} \sin x \, dx
\]
Calculating each part:
1. \(\int \cos x \, dx = \sin x\)
2. \(\int \sin x \, dx = -\cos x\)
Now applying the limits:
\[
\left[ \sin x \right]_{0}^{\frac{\pi}{4}} - \left[ -\cos x \right]_{0}^{\frac{\pi}{4}} = \left( \sin \frac{\pi}{4} - \sin 0 \right) - \left( -\cos \frac{\pi}{4} + \cos 0 \right)
\]
Calculating the values:
\[
\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \sin 0 = 0, \quad \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \cos 0 = 1
\]
Thus, we have:
\[
\frac{1}{\sqrt{2}} - 0 - \left( -\frac{1}{\sqrt{2}} + 1 \right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1 = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1
\]
### Step 2: Evaluate the second integral
\[
\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx
\]
Again, we can split it:
\[
\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \sin x \, dx - \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos x \, dx
\]
Calculating each part:
1. \(\int \sin x \, dx = -\cos x\)
2. \(\int \cos x \, dx = \sin x\)
Now applying the limits:
\[
\left[ -\cos x \right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} - \left[ \sin x \right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} = \left( -\cos \frac{5\pi}{4} + \cos \frac{\pi}{4} \right) - \left( \sin \frac{5\pi}{4} - \sin \frac{\pi}{4} \right)
\]
Calculating the values:
\[
\cos \frac{5\pi}{4} = -\frac{1}{\sqrt{2}}, \quad \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \sin \frac{5\pi}{4} = -\frac{1}{\sqrt{2}}, \quad \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}
\]
Thus, we have:
\[
\left( -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} \right) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \left( -\frac{2}{\sqrt{2}} \right) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}
\]
### Step 3: Evaluate the third integral
\[
\int_{2\pi}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx
\]
This integral can be rewritten by reversing the limits:
\[
-\int_{\frac{\pi}{4}}^{2\pi} (\cos x - \sin x) \, dx
\]
Calculating:
\[
-\left( \int_{\frac{\pi}{4}}^{2\pi} \cos x \, dx - \int_{\frac{\pi}{4}}^{2\pi} \sin x \, dx \right)
\]
Calculating each part:
1. \(\int \cos x \, dx = \sin x\)
2. \(\int \sin x \, dx = -\cos x\)
Now applying the limits:
\[
-\left( \left[ \sin x \right]_{\frac{\pi}{4}}^{2\pi} - \left[ -\cos x \right]_{\frac{\pi}{4}}^{2\pi} \right) = -\left( \left( \sin 2\pi - \sin \frac{\pi}{4} \right) - \left( -\cos 2\pi + \cos \frac{\pi}{4} \right) \right)
\]
Calculating the values:
\[
\sin 2\pi = 0, \quad \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \cos 2\pi = 1, \quad \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}
\]
Thus, we have:
\[
-\left( 0 - \frac{1}{\sqrt{2}} - ( -1 + \frac{1}{\sqrt{2}} ) \right) = -\left( -\frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} \right) = -\left( 1 - \frac{2}{\sqrt{2}} \right) = -1 + \sqrt{2}
\]
### Step 4: Combine all results
Now we combine all three integrals:
\[
(\sqrt{2} - 1) + (2\sqrt{2}) + (-1 + \sqrt{2})
\]
Combining like terms:
\[
\sqrt{2} + 2\sqrt{2} + \sqrt{2} - 1 - 1 = 4\sqrt{2} - 2
\]
### Final Answer
Thus, the final result is:
\[
\boxed{4\sqrt{2} - 2}
\]
