Primitive of Cube of Secant Function/Proof 1

Theorem

$\ds \int \sec^3 x \rd x = \frac 1 2 \paren {\sec x \tan x + \ln \size {\sec x + \tan x} } + C$

Proof

From Primitive of $\sec^3 a x$:

$\ds \int \sec^3 a x \rd x = \frac 1 {2 a} \paren {\sec a x \tan a x + \ln \size {\sec a x + \tan a x} } + C$

The result follows on setting $a = 1$.

$\blacksquare$