The method of integration by parts is used to find the integral
\[ \int \sec^3 x \; dx \]
The integration by parts is expressed as follows
\[ \int u' v \; dx = u v - \int u v' \; dx \]
Let \( v = \sec x \) and \( u' = \sec^2 x \); hence \( u = \displaystyle \int \sec^2 x \; dx = \tan x \) and \( v' = \sec x \tan x \) to write
\[ \int \sec^3 x \; dx = \int \sec^2 x \; \sec x \; dx \\ = \tan x \; \sec x - \int \tan x \; \sec x \; \tan x \; dx \\ = \tan x \; \sec x - \int \tan^2 x \; \sec x \; dx \qquad (I)\]
Use the identity \( \tan^2 x = \sec^2 x - 1 \) to write the last integral as follows
\[ \int \tan^2 x \; \sec x \; dx = \int (\sec^2 x - 1) \sec x dx \\
= \int \sec^3 x \; dx - \int \sec x \; dx \]
Substitute in (I) and write the given integral as follows
\[ \int \sec^3 x \; dx = \tan x \; \sec x + \int \sec x \; dx - \int \sec^3 x \; dx \]
Add \( \displaystyle \int \sec^3 x \; dx \) to both sides of the above equation and simplify to obtain
\[ 2 \int \sec^3 x \; dx = \tan x \; \sec x + \int \sec x \; dx \]
Use the common integral \( \displaystyle \int \sec x \; dx = \ln |\tan x + sec x| \) to write the above as
\[ 2 \int \sec^3 x \; dx = \tan x \; \sec x + \ln |\tan x + sec x| \]
Divide all terms by \( 2 \) to obtain the final answer.
\[ \boxed { \int \sec^3 x \; dx = \dfrac{1}{2} \left( \tan x \; \sec x + \ln |\tan x + \sec x| \right) + c } \]
