Primitive of Secant Function/Secant plus Tangent Form/Proof 1

Theorem

$\ds \int \sec x \rd x = \ln \size {\sec x + \tan x} + C$

where $\sec x + \tan x \ne 0$.

Let:

$\ds u$

$=$

$\ds \tan x + \sec x$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac \d {\d x} \tan x + \frac \d {\d x} \sec x$

$\ds $

$=$

$\ds \sec^2 x + \frac \d {\d x} \sec x$

$\ds $

$=$

$\ds \sec^2 x + \sec x \tan x$

$\ds $

$=$

$\ds \sec x \paren {\sec x + \tan x}$

factorising

Then:

$\ds \int \sec x \rd x$

$=$

$\ds \int \frac {\sec x \paren {\sec x + \tan x} } {\sec x + \tan x} \rd x$

multiplying top and bottom by $\sec x + \tan x$

$\ds $

$=$

$\ds \ln \size {\sec x + \tan x} + C$

$\blacksquare$