1. **State the problem:** Evaluate the definite integral $$\int_0^{\frac{\pi}{8}} \frac{5}{\cos^2\left(\frac{\pi}{4} - 2x\right)} \, dx$$ 2. **Recall the formula and rules:** We know that $$\frac{1}{\cos^2 u} = \sec^2 u$$ and the integral of $\sec^2 u$ with respect to $u$ is $$\int \sec^2 u \, du = \tan u + C$$ 3. **Substitution:** Let $$u = \frac{\pi}{4} - 2x$$ Then, $$\frac{du}{dx} = -2 \implies dx = \frac{du}{-2}$$ 4. **Change the limits:** When $x=0$, $$u = \frac{\pi}{4} - 2 \cdot 0 = \frac{\pi}{4}$$ When $x=\frac{\pi}{8}$, $$u = \frac{\pi}{4} - 2 \cdot \frac{\pi}{8} = \frac{\pi}{4} - \frac{\pi}{4} = 0$$ 5. **Rewrite the integral:** $$\int_0^{\frac{\pi}{8}} \frac{5}{\cos^2\left(\frac{\pi}{4} - 2x\right)} \, dx = \int_{u=\frac{\pi}{4}}^{u=0} 5 \sec^2 u \cdot \frac{du}{-2} = -\frac{5}{2} \int_{\frac{\pi}{4}}^{0} \sec^2 u \, du$$ 6. **Flip the limits to remove the negative sign:** $$-\frac{5}{2} \int_{\frac{\pi}{4}}^{0} \sec^2 u \, du = \frac{5}{2} \int_0^{\frac{\pi}{4}} \sec^2 u \, du$$ 7. **Integrate:** $$\frac{5}{2} \int_0^{\frac{\pi}{4}} \sec^2 u \, du = \frac{5}{2} \left[ \tan u \right]_0^{\frac{\pi}{4}} = \frac{5}{2} (\tan \frac{\pi}{4} - \tan 0)$$ 8. **Evaluate the tangent values:** $$\tan \frac{\pi}{4} = 1, \quad \tan 0 = 0$$ So, $$\frac{5}{2} (1 - 0) = \frac{5}{2}$$ **Final answer:** $$\boxed{\frac{5}{2}}$$