1. **State the problem:** Evaluate the integral $$\int \tan^4 x \sec x \, dx$$. 2. **Recall relevant identities and formulas:** - We know that $$\frac{d}{dx}(\sec x) = \sec x \tan x$$. - Also, $$\tan^2 x = \sec^2 x - 1$$. 3. **Rewrite the integrand:** Express $$\tan^4 x$$ in terms of $$\sec x$$: $$\tan^4 x = (\tan^2 x)^2 = (\sec^2 x - 1)^2 = \sec^4 x - 2 \sec^2 x + 1$$. So the integral becomes: $$\int \tan^4 x \sec x \, dx = \int (\sec^4 x - 2 \sec^2 x + 1) \sec x \, dx = \int (\sec^5 x - 2 \sec^3 x + \sec x) \, dx$$. 4. **Split the integral:** $$\int \sec^5 x \, dx - 2 \int \sec^3 x \, dx + \int \sec x \, dx$$. 5. **Use reduction formulas for powers of secant:** - The integral of $$\sec x$$ is: $$\int \sec x \, dx = \ln |\sec x + \tan x| + C$$. - The integral of $$\sec^3 x$$ is: $$\int \sec^3 x \, dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| + C$$. - For $$\int \sec^5 x \, dx$$, use the reduction formula: $$\int \sec^n x \, dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} x \, dx$$. Applying for $$n=5$$: $$\int \sec^5 x \, dx = \frac{\sec^3 x \tan x}{4} + \frac{3}{4} \int \sec^3 x \, dx$$. 6. **Substitute the known integral for $$\int \sec^3 x \, dx$$:** $$\int \sec^5 x \, dx = \frac{\sec^3 x \tan x}{4} + \frac{3}{4} \left( \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| \right) + C$$ Simplify: $$= \frac{\sec^3 x \tan x}{4} + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln |\sec x + \tan x| + C$$. 7. **Now substitute back into the original integral:** $$\int \tan^4 x \sec x \, dx = \left( \frac{\sec^3 x \tan x}{4} + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln |\sec x + \tan x| \right) - 2 \left( \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| \right) + \ln |\sec x + \tan x| + C$$ Simplify the 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: $$\frac{\sec^3 x \tan x}{4} + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln |\sec x + \tan x| - \sec x \tan x - \ln |\sec x + \tan x| + \ln |\sec x + \tan x| + C$$ Notice $$- \ln |\sec x + \tan x| + \ln |\sec x + \tan x| = 0$$, so they cancel out. Combine $$\sec x \tan x$$ terms: $$\frac{3}{8} \sec x \tan x - \sec x \tan x = \frac{3}{8} \sec x \tan x - \frac{8}{8} \sec x \tan x = -\frac{5}{8} \sec x \tan x$$. 8. **Final answer:** $$\int \tan^4 x \sec x \, dx = \frac{\tan x \sec^3 x}{4} - \frac{5}{8} \tan x \sec x + \frac{3}{8} \ln |\sec x + \tan x| + C$$. This matches option (a).