Step 1: Rewrite the integral and apply integration by parts. Let $I = \int \sec^3 x \, dx$. We can write $\sec^3 x = \sec x \cdot \sec^2 x$. Use integration by parts, $\int u \, dv = uv - \int v \, du$. Let $u = \sec x$ and $dv = \sec^2 x \, dx$. Then $du = \sec x \tan x \, dx$ and $v = \tan x$. $$I = \sec x \tan x - \int \tan x (\sec x \tan x) \, dx$$ $$I = \sec x \tan x - \int \sec x \tan^2 x \, dx$$ Step 2: Use the trigonometric identity $\tan^2 x = \sec^2 x - 1$. Substitute this identity into the integral: $$I = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx$$ $$I = \sec x \tan x - \int (\sec^3 x - \sec x) \, dx$$ $$I = \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx$$ Step 3: Solve for $I$. Notice that the original integral $I = \int \sec^3 x \, dx$ appears on the right side. $$I = \sec x \tan x - I + \int \sec x \, dx$$ Add $I$ to both sides: $$2I = \sec x \tan x + \int \sec x \, dx$$ Step 4: Evaluate the integral of $\sec x$. The integral of $\sec x$ is $\ln|\sec x + \tan x|$. $$2I = \sec x \tan x + \ln|\sec x + \tan x| + C'$$ Divide by 2 to solve for $I$: $$I = \frac{1}{2} (\sec x \tan x + \ln|\sec x + \tan x|) + C$$ where $C = \frac{C'}{2}$ is the constant of integration. The final answer is $\boxed{\frac{1}{2} (\sec x \tan x + \ln|\sec x + \tan x|) + C}$.