# Primitive of Secant Function/Secant plus Tangent Form

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## Theorem

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

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

## Proof 1

Let:

 $\ds u$ $=$ $\ds \tan x + \sec x$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds \frac \d {\d x} \tan x + \frac \d {\d x} \sec x$ Linear Combination of Derivatives $\ds$ $=$ $\ds \sec^2 x + \frac \d {\d x} \sec x$ Derivative of Tangent Function $\ds$ $=$ $\ds \sec^2 x + \sec x \tan x$ Derivative of Secant Function $\ds$ $=$ $\ds \sec x \paren {\sec x + \tan x}$ 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 $\sec x + \tan x$ $\ds$ $=$ $\ds \ln \size {\sec x + \tan x} + C$ Primitive of Function under its Derivative

$\blacksquare$

## Proof 2

 $\ds \int \sec x \rd x$ $=$ $\ds \int \frac 1 {\cos x} \rd x$ Secant is Reciprocal of Cosine

We make the Weierstrass Substitution:

 $\ds u$ $=$ $\ds \tan \frac x 2$ $\ds \leadsto \ \$ $\ds \cos x$ $=$ $\ds \frac {1 - u^2} {1 + u^2}$ $\ds \leadsto \ \$ $\ds \frac {\d x} {\d u}$ $=$ $\ds \frac 2 {u^2 + 1}$ $\ds \leadsto \ \$ $\ds \int \frac 1 {\cos x} \rd x$ $=$ $\ds \int \frac {1 + u^2} {1 - u^2} \frac 2 {u^2 + 1} \rd u$ $\ds$ $=$ $\ds 2 \int \frac 1 {1 - u^2} \rd u$ $\ds$ $=$ $\ds \ln \size {\frac {1 + u} {1 - u} } + C$ Primitive of $\dfrac 1 {a^2 - u^2}$: Logarithm Form $\ds$ $=$ $\ds \ln \size {\frac {1 + \tan \frac x 2} {1 - \tan \frac x 2} } + C$ $\ds$ $=$ $\ds \ln \size {\sec x + \tan x} + C$ One Plus Tangent Half Angle over One Minus Tangent Half Angle

$\blacksquare$