Primitive of Cube of Tangent of x

Theorem

$\ds \int \tan^3 x \rd x$

$=$

$\ds \frac {\tan^2 x} 2 + \ln \size {\cos x} + C$

$\ds $

$=$

$\ds \frac {\tan^2 x} 2 - \ln \size {\sec x} + C$

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

The result follows on setting $a = 1$, then using Primitive of $\tan x$: Secant Form

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $30$.