1. **State the problem:** Evaluate the integral $$\int 12 \tan^3(x) \, dx$$. 2. **Recall the formula and rules:** We use the identity $$\tan^3(x) = \tan(x) \cdot \tan^2(x)$$ and the Pythagorean identity $$\tan^2(x) = \sec^2(x) - 1$$. 3. **Rewrite the integral:** $$\int 12 \tan^3(x) \, dx = \int 12 \tan(x) (\sec^2(x) - 1) \, dx = 12 \int \tan(x) \sec^2(x) \, dx - 12 \int \tan(x) \, dx$$ 4. **Evaluate each integral separately:** - For $$\int \tan(x) \sec^2(x) \, dx$$, let $$u = \tan(x)$$, so $$du = \sec^2(x) \, dx$$. Thus, $$\int \tan(x) \sec^2(x) \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{\tan^2(x)}{2} + C$$. - For $$\int \tan(x) \, dx$$, recall that $$\int \tan(x) \, dx = -\ln|\cos(x)| + C$$. 5. **Combine results:** $$12 \int \tan(x) \sec^2(x) \, dx - 12 \int \tan(x) \, dx = 12 \cdot \frac{\tan^2(x)}{2} - 12 (-\ln|\cos(x)|) + C = 6 \tan^2(x) + 12 \ln|\cos(x)| + C$$ 6. **Final answer:** $$\int 12 \tan^3(x) \, dx = 6 \tan^2(x) + 12 \ln|\cos(x)| + C$$