1. **State the problem:** Calculate the definite integral $$\int_0^{\frac{\pi}{4}} \tan^8(\theta) \sec^2(\theta) \, d\theta$$. 2. **Identify substitution:** Notice that the derivative of $\tan(\theta)$ is $\sec^2(\theta)$, which appears in the integrand. This suggests the substitution: $$u = \tan(\theta) \implies du = \sec^2(\theta) \, d\theta$$ 3. **Change limits of integration:** When $\theta = 0$, $u = \tan(0) = 0$. When $\theta = \frac{\pi}{4}$, $u = \tan\left(\frac{\pi}{4}\right) = 1$. 4. **Rewrite the integral:** Substituting, the integral becomes: $$\int_0^1 u^8 \, du$$ 5. **Integrate:** Use the power rule for integration: $$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$ So, $$\int_0^1 u^8 \, du = \left[ \frac{u^9}{9} \right]_0^1 = \frac{1^9}{9} - \frac{0^9}{9} = \frac{1}{9}$$ 6. **Final answer:** $$\boxed{\frac{1}{9}}$$