# Primitive of x by Square of Secant of a x

## Theorem

$\displaystyle \int x \sec^2 a x \ \mathrm d x = \frac {x \tan a x} a + \frac 1 {a^2} \ln \left\vert{\cos a x}\right\vert + C$

## Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

 $\displaystyle u$ $=$ $\displaystyle x$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle 1$ Derivative of Identity Function

and let:

 $\displaystyle \frac {\mathrm d v} {\mathrm d x}$ $=$ $\displaystyle \sec^2 a x$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {\tan a x} a$ Primitive of $\sec^2 a x$

Then:

 $\displaystyle \int x \sec^2 a x \ \mathrm d x$ $=$ $\displaystyle \frac {x \tan a x} a - \int \frac {\tan a x} a \ \mathrm d x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {x \tan a x} a - \frac 1 a \left({\frac {-1} a \ln \left\vert{\cos a x}\right\vert}\right) + C$ Primitive of $\tan a x$ $\displaystyle$ $=$ $\displaystyle \frac {x \tan a x} a + \frac 1 {a^2} \ln \left\vert{\cos a x}\right\vert + C$ simplifying

$\blacksquare$