Primitive of Cube of Secant of a x/Proof 1

Theorem

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

Proof

\(\ds \int \sec^3 x \rd x\)

\(=\)

\(\ds \frac {\sec a x \tan a x} {2 a} + \frac 1 2 \int \sec a x \rd x\)

Primitive of $\sec^n a x$ where $n = 3$

\(\ds \)

\(=\)

\(\ds \frac {\sec a x \tan a x} {2 a} + \frac 1 2 \paren {\frac 1 a \ln \size {\sec a x + \tan a x} }\)

Primitive of $\sec a x$

\(\ds \)

\(=\)

\(\ds \frac 1 {2 a} \paren {\sec a x \tan a x + \ln \size {\sec a x + \tan a x} } + C\)

simplifying

$\blacksquare$