# Primitive of Reciprocal of x by Root of x squared minus a squared/Arcsecant Form

## Theorem

$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \arcsec \size {\frac x a} + C$

for $0 < a < \size x$.

## Proof

We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:

$x > a$

or:

$x < -a$

where it is assumed that $a > 0$.

Consider the arcsecant substitution:

$u = \arcsec {\dfrac x a}$

which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$.

That is:

$\size x \ge a$

and it is seen that $u = \arcsec {\dfrac x a}$ is defined over the whole domain of the integrand.

Hence:

 $\ds u$ $=$ $\ds \arcsec {\frac x a}$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds a \sec u$ $\ds \leadsto \ \$ $\ds \frac {\d x} {\d u}$ $=$ $\ds a \sec u \tan u$ Derivative of Secant Function

Let $x > a$.

 $\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} }$ $=$ $\ds \int \frac {a \sec u \tan u} {a \sec u \sqrt {a^2 \sec^2 u - a^2} } \rd u$ Integration by Substitution $\ds$ $=$ $\ds \frac a {a^2} \int \frac {\sec u \tan u} {\sec u \sqrt {\sec^2 u - 1} } \rd u$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds \frac 1 a \int \frac {\sec u \tan u} {\sec u \tan u} \rd u$ Difference of Squares of Secant and Tangent $\ds$ $=$ $\ds \frac 1 a \int 1 \rd u$ $\ds$ $=$ $\ds \frac 1 a u + C$ Integral of Constant $\ds$ $=$ $\ds \frac 1 a \arcsec {\frac x a} + C$ Definition of $u$ $\ds$ $=$ $\ds \frac 1 a \arcsec {\frac {\size x} a} + C$ Definition of Absolute Value: $\size x = x$ for $x > 0$

Now suppose $x < -a$.

Let $z = -x$.

Then:

$\d x = -\d z$

and we then have:

 $\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} }$ $=$ $\ds \int \frac {-\d z} {\paren {-z} \sqrt {\paren {-z}^2 - a^2} }$ Integration by Substitution $\ds$ $=$ $\ds \int \frac {\d z} {z \sqrt {z^2 - a^2} }$ simplifying $\ds$ $=$ $\ds \frac 1 a \arcsec {\frac z a} + C$ from above $\ds$ $=$ $\ds \frac 1 a \arcsec {\frac {\paren {-x} } a} + C$ substituting back for $x$ $\ds$ $=$ $\ds \frac 1 a \arcsec {\frac {\size x} a} + C$ Definition of Absolute Value: $\size x = -x$ for $x < 0$

The result follows.

$\blacksquare$