\section{Primitive of Inverse Hyperbolic Secant of x over a}
Tags: Primitives involving Inverse Hyperbolic Secant Function, Primitive of Inverse Hyperbolic Secant of x over a

\begin{theorem}
:$\ds \int \arsech \frac x a \rd x = x \arsech \dfrac x a + a \arcsin \dfrac x a + C$
where $\arsech$ denotes the real area hyperbolic secant.
\end{theorem}

\begin{proof}
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arsech \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {-a} {x \sqrt {a^2 - x^2} }
      | c = Derivative of $\arsech \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = x
      | c = Primitive of Constant
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \arsech \frac x a \rd x
      | r = x \arsech \frac x a - \int x \paren {\frac {-a} {x \sqrt {a^2 - x^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = x \arsech \frac x a + a \int \frac {\d x} {\sqrt {a^2 - x^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = x \arsech \frac x a + \arcsin \frac x a + C
      | c = Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$
}}
{{end-eqn}}
{{qed}}
\end{proof}

