1. **State the problem:** We want to evaluate the integral $$\int 9 \sec^4 x \tan^8 x \, dx.$$\n\n2. **Recall formulas and identities:** We use the identity $$\sec^2 x = 1 + \tan^2 x$$ and the derivative $$\frac{d}{dx}(\tan x) = \sec^2 x.$$\n\n3. **Rewrite the integral:** Express powers of secant in terms of $$\sec^2 x$$ to use substitution. Write $$\sec^4 x = (\sec^2 x)^2$$ so the integral becomes $$9 \int \sec^4 x \tan^8 x \, dx = 9 \int (\sec^2 x)^2 \tan^8 x \, dx.$$\n\n4. **Use substitution:** Let $$u = \tan x,$$ then $$du = \sec^2 x \, dx.$$\n\n5. **Rewrite the integral in terms of $$u$$:** We have $$\sec^2 x \, dx = du,$$ so $$\sec^4 x \, dx = \sec^2 x (\sec^2 x \, dx) = \sec^2 x \, du.$$\n\nRewrite the integral as $$9 \int \sec^2 x \tan^8 x \, du = 9 \int \sec^2 x u^8 \, du.$$\n\n6. **Express $$\sec^2 x$$ in terms of $$u$$:** Since $$u = \tan x,$$ $$\sec^2 x = 1 + u^2.$$\n\n7. **Substitute:** The integral becomes $$9 \int (1 + u^2) u^8 \, du = 9 \int (u^8 + u^{10}) \, du.$$\n\n8. **Integrate term-by-term:** $$9 \left( \frac{u^9}{9} + \frac{u^{11}}{11} \right) + C = u^9 + \frac{9}{11} u^{11} + C.$$\n\n9. **Back-substitute $$u = \tan x$$:** $$\tan^9 x + \frac{9}{11} \tan^{11} x + C.$$\n\n**Final answer:** $$\int 9 \sec^4 x \tan^8 x \, dx = \tan^9 x + \frac{9}{11} \tan^{11} x + C.$$