# Integral of $\sec^3(x)$

 

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. $\boxed { \int \sec^3 x \; dx = \dfrac{1}{2} \left( \tan x \; \sec x + \ln |\tan x + \sec x| \right) + c }$