# Primitive of Power of Tangent of a x by Square of Secant of a x

## Theorem

$\displaystyle \int \tan^n a x \sec^2 a x \rd x = \frac {\tan^{n + 1} a x} {\paren {n + 1} a} + C$

## Proof

 $\displaystyle z$ $=$ $\displaystyle \tan a x$ $\displaystyle \implies \ \$ $\displaystyle \frac {\d z} {\d x}$ $=$ $\displaystyle a \sec^2 a x$ Derivative of Tangent Function: Corollary $\displaystyle \implies \ \$ $\displaystyle \int \tan^n a x \sec^2 a x \rd x$ $=$ $\displaystyle \int \frac 1 a z^n \rd z$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 a \frac {z^{n + 1} } {n + 1}$ Primitive of Power $\displaystyle$ $=$ $\displaystyle \frac {\tan^{n + 1} a x} {\paren {n + 1} a} + C$ substituting for $z$ and simplifying

$\blacksquare$