1. The problem is to find the integral of $\tan^6(x)$ with respect to $x$. 2. We use the identity $\tan^2(x) = \sec^2(x) - 1$ to rewrite powers of tangent in terms of secant. 3. Express $\tan^6(x)$ as $(\tan^2(x))^3 = (\sec^2(x) - 1)^3$. 4. Expand the cube: $$(\sec^2(x) - 1)^3 = \sec^6(x) - 3\sec^4(x) + 3\sec^2(x) - 1$$ 5. So the integral becomes $$\int \tan^6(x) dx = \int (\sec^6(x) - 3\sec^4(x) + 3\sec^2(x) - 1) dx$$ 6. Integrate term by term: - $\int \sec^6(x) dx$ - $-3 \int \sec^4(x) dx$ - $+3 \int \sec^2(x) dx$ - $- \int 1 dx$ 7. Use reduction formulas or known integrals: - $\int \sec^2(x) dx = \tan(x) + C$ - $\int 1 dx = x + C$ - For $\int \sec^4(x) dx$, use $\sec^4(x) = (\sec^2(x))^2$ and reduction formulas: $$\int \sec^4(x) dx = \frac{\sec^2(x) \tan(x)}{3} + \frac{2}{3} \int \sec^2(x) dx = \frac{\sec^2(x) \tan(x)}{3} + \frac{2}{3} \tan(x) + C$$ - For $\int \sec^6(x) dx$, use reduction formula: $$\int \sec^6(x) dx = \frac{\sec^4(x) \tan(x)}{5} + \frac{4}{5} \int \sec^4(x) dx$$ 8. Substitute $\int \sec^4(x) dx$ into the above: $$\int \sec^6(x) dx = \frac{\sec^4(x) \tan(x)}{5} + \frac{4}{5} \left( \frac{\sec^2(x) \tan(x)}{3} + \frac{2}{3} \tan(x) \right) + C$$ 9. Simplify: $$= \frac{\sec^4(x) \tan(x)}{5} + \frac{4 \sec^2(x) \tan(x)}{15} + \frac{8 \tan(x)}{15} + C$$ 10. Now combine all integrals: $$\int \tan^6(x) dx = \left( \frac{\sec^4(x) \tan(x)}{5} + \frac{4 \sec^2(x) \tan(x)}{15} + \frac{8 \tan(x)}{15} \right) - 3 \left( \frac{\sec^2(x) \tan(x)}{3} + \frac{2}{3} \tan(x) \right) + 3 \tan(x) - x + C$$ 11. Simplify the terms: $$= \frac{\sec^4(x) \tan(x)}{5} + \frac{4 \sec^2(x) \tan(x)}{15} + \frac{8 \tan(x)}{15} - \sec^2(x) \tan(x) - 2 \tan(x) + 3 \tan(x) - x + C$$ 12. Combine like terms: $$\frac{4}{15} \sec^2(x) \tan(x) - \sec^2(x) \tan(x) = \left( \frac{4}{15} - 1 \right) \sec^2(x) \tan(x) = -\frac{11}{15} \sec^2(x) \tan(x)$$ $$\frac{8}{15} \tan(x) - 2 \tan(x) + 3 \tan(x) = \left( \frac{8}{15} - 2 + 3 \right) \tan(x) = \frac{23}{15} \tan(x)$$ 13. Final answer: $$\int \tan^6(x) dx = \frac{\sec^4(x) \tan(x)}{5} - \frac{11}{15} \sec^2(x) \tan(x) + \frac{23}{15} \tan(x) - x + C$$