# Primitive of Inverse Hyperbolic Secant of x over a

## Theorem

$\displaystyle \int \sech^{-1} \frac x a \rd x = \begin{cases} x \sech^{-1} \dfrac x a + a \arcsin \dfrac x a + C & : \sech^{-1} \dfrac x a > 0 \\ x \sech^{-1} \dfrac x a - a \arcsin \dfrac x a + C & : \sech^{-1} \dfrac x a < 0 \end{cases}$

## Proof

With a view to expressing the primitive in the form:

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

let:

 $\displaystyle u$ $=$ $\displaystyle \sech^{-1} \frac x a$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d u} {\d x}$ $=$ $\displaystyle \frac {-a} {x \sqrt {a^2 - x^2} }$ Derivative of $\sech^{-1} \dfrac x a$

and let:

 $\displaystyle \frac {\d v} {\d x}$ $=$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle v$ $=$ $\displaystyle x$ Primitive of Constant

Then:

 $\displaystyle \int \sech^{-1} \frac x a \rd x$ $=$ $\displaystyle x \sech^{-1} \frac x a - \int x \paren {\frac {-a} {x \sqrt {a^2 - x^2} } } \rd x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle x \sech^{-1} \frac x a + a \int \frac {\d x} {\sqrt {a^2 - x^2} } + C$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle x \sech^{-1} \frac x a + \arcsin \frac x a + C$ Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$