1. **State the problem:** Evaluate the integral $$\int \tan^4 x \sec x \, dx$$. 2. **Recall relevant identities:** - $$\frac{d}{dx}(\sec x) = \sec x \tan x$$ - $$\tan^2 x = \sec^2 x - 1$$ 3. **Rewrite the integrand:** $$\tan^4 x = (\tan^2 x)^2 = (\sec^2 x - 1)^2 = \sec^4 x - 2 \sec^2 x + 1$$ So, $$\int \tan^4 x \sec x \, dx = \int (\sec^4 x - 2 \sec^2 x + 1) \sec x \, dx = \int (\sec^5 x - 2 \sec^3 x + \sec x) \, dx$$ 4. **Split the integral:** $$\int \sec^5 x \, dx - 2 \int \sec^3 x \, dx + \int \sec x \, dx$$ 5. **Use known integrals and reduction formulas:** - $$\int \sec x \, dx = \ln |\sec x + \tan x| + C$$ - $$\int \sec^3 x \, dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| + C$$ - Reduction formula for $$\int \sec^n x \, dx$$: $$\int \sec^n x \, dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} x \, dx$$ 6. **Apply reduction formula for $$n=5$$:** $$\int \sec^5 x \, dx = \frac{\sec^3 x \tan x}{4} + \frac{3}{4} \int \sec^3 x \, dx$$ Substitute $$\int \sec^3 x \, dx$$: $$= \frac{\sec^3 x \tan x}{4} + \frac{3}{4} \left( \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| \right) + C$$ 7. **Simplify:** $$= \frac{\sec^3 x \tan x}{4} + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln |\sec x + \tan x| + C$$ 8. **Substitute back into original integral:** $$\int \tan^4 x \sec x \, dx = \left( \frac{\sec^3 x \tan x}{4} + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln |\sec x + \tan x| \right) - 2 \left( \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| \right) + \ln |\sec x + \tan x| + C$$ 9. **Simplify terms:** - $$-2 \times \frac{1}{2} \sec x \tan x = - \sec x \tan x$$ - $$-2 \times \frac{1}{2} \ln |\sec x + \tan x| = - \ln |\sec x + \tan x|$$ So the integral becomes: $$\frac{\sec^3 x \tan x}{4} + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln |\sec x + \tan x| - \sec x \tan x - \ln |\sec x + \tan x| + \ln |\sec x + \tan x| + C$$ 10. **Cancel and combine:** - The logarithm terms $$- \ln |\sec x + \tan x| + \ln |\sec x + \tan x| = 0$$ - Combine $$\sec x \tan x$$ terms: $$\frac{3}{8} \sec x \tan x - \sec x \tan x = \frac{3}{8} \sec x \tan x - \frac{8}{8} \sec x \tan x = -\frac{5}{8} \sec x \tan x$$ **Final answer:** $$\int \tan^4 x \sec x \, dx = \frac{\tan x \sec^3 x}{4} - \frac{5}{8} \tan x \sec x + \frac{3}{8} \ln |\sec x + \tan x| + C$$