# Primitive of Reciprocal of x by Root of a squared minus x squared/Inverse Hyperbolic Secant Form

## Theorem

For $a > 0$ and $0 < \size x < a$:

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

## Proof

We note that $\sech^{-1} \dfrac x a$ is defined for $x \in \hointl 0 a$.

Hence we treat the two cases $x > 0$ and $x < 0$ separately.

First let $x > 0$.

Note that $\dfrac 1 {x \sqrt {a^2 - x^2} }$ is not defined at $\pm a$, so we are concerned only about the interval $\openint 0 a$.

Then:

 $\ds u$ $=$ $\ds \sech^{-1} {\frac x a}$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds a \sech u$ $\ds \leadsto \ \$ $\ds \frac {\d x} {\d u}$ $=$ $\ds -a \sech u \tanh u$ Derivative of Hyperbolic Secant $\ds \leadsto \ \$ $\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} }$ $=$ $\ds \int \frac {-a \sech u \tanh u} {a \sech u \sqrt {a^2 - a^2 \sech^2 u} } \rd u$ Integration by Substitution $\ds$ $=$ $\ds -\frac a {a^2} \int \frac {\sech u \tanh u} {\sech u \sqrt {1 - \sech^2 u} } \rd u$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds -\frac 1 a \int \frac {\sech u \tanh u} {\sech u \tanh u} \rd u$ Sum of Squares of Hyperbolic Secant and Tangent $\ds$ $=$ $\ds -\frac 1 a \int 1 \rd u$ $\ds$ $=$ $\ds -\frac 1 a u + C$ Integral of Constant $\ds$ $=$ $\ds -\frac 1 a \sech^{-1} {\frac x a} + C$ Definition of $u$

$\Box$

Now let $x < 0$.

Let $z = -x$.

Then:

 $\ds z$ $=$ $\ds -x$ $\ds \leadsto \ \$ $\ds \frac {\d z} {\d x}$ $=$ $\ds -1$ Derivative of Constant Multiple $\ds \leadsto \ \$ $\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} }$ $=$ $\ds \int \frac {-\d z} {\paren {-z} \sqrt {a^2 - z^2} }$ Integration by Substitution $\ds$ $=$ $\ds \int \frac {\d z} {z \sqrt {a^2 - z^2} }$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds -\frac 1 a \sech^{-1} {\frac z a} + C$ from above: $z \in \hointl 0 a$ $\ds$ $=$ $\ds -\frac 1 a \sech^{-1} {\frac {-x} a} + C$ Definition of $z$

$\Box$

So when $x > 0$:

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

and when $x < 0$:

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

It follows that for $\size x \in \openint 0 a$:

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

by definition of absolute value.

$\blacksquare$