Evaluate; integral of sec³x dx

Step 1: Use integration by parts. Let $u = \sec x$ and $dv = \sec^2 x \, dx$. Then $du = \sec x \tan x \, dx$ and $v = \tan x$.

Step 2: Apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$. $\int \sec^3 x \, dx = \sec x \tan x - \int \tan x (\sec x \tan x) \, dx$ $\int \sec^3 x \, dx = \sec x \tan x - \int \sec x \tan^2 x \, dx$

Step 3: Use the identity $\tan^2 x = \sec^2 x - 1$. $\int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx$ $\int \sec^3 x \, dx = \sec x \tan x - \int (\sec^3 x - \sec x) \, dx$

Step 4: Distribute the integral. $\int \sec^3 x \, dx = \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx$

Step 5: Let $I = \int \sec^3 x \, dx$. The equation becomes: $I = \sec x \tan x - I + \int \sec x \, dx$

Step 6: Add $I$ to both sides. $2I = \sec x \tan x + \int \sec x \, dx$

Step 7: Evaluate the integral of $\sec x$. $\int \sec x \, dx = \ln|\sec x + \tan x| + C$

Step 8: Substitute this back into the equation for $2I$. $2I = \sec x \tan x + \ln|\sec x + \tan x| + C$

Step 9: Solve for $I$. $I = \frac{1}{2} (\sec x \tan x + \ln|\sec x + \tan x|) + C$

The final answer is $\boxed{\frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) + C}$.