Primitive of Secant Function/Secant plus Tangent Form

From ProofWiki
Jump to navigation Jump to search

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

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Secant_Function/Secant_plus_Tangent_Form&oldid=625661"