A calculus of the absurd

16.2.3 Inverse function of \(\sinh (x)\)

Taking \(\sinh (x)\), as an example, we know that

\begin{equation} \sinh (x) = \frac {e^x-e^{-x}}{2} \end{equation}

We can write \(x\) in terms of \(\sinh (x)\), by multiplying the fraction through by \(\frac {e^{x}}{e^{x}}\).

\begin{equation} \sinh (x) = \frac {e^{2x}-1}{2e^{x}} \end{equation}

Here we can substitute \(u=e^{x}\)

\begin{equation} \sinh (x) = \frac {u^2-1}{2u} \end{equation}

And then we can rearrange this to give a quadratic in terms of \(u\).

\begin{equation} u^2 - 2u\sinh (x) - 1 = 0 \end{equation}

By applying the quadratic formula, we can solve this for \(u\)

\begin{equation} u = \frac { -(-2\sinh (x)) \pm \sqrt {(-2\sinh (x))^2 - 4(1)(-1)} } {2} \end{equation}

We can then simplify this a bit

\begin{equation} u = \frac { 2\sinh (x) \pm \sqrt {4}\sqrt {\sinh (x)^2 + 1} } {2} \end{equation}

If we then reverse the substitution103103 Arguably, the substitution didn’t help here, but substitutions do often make it easier to spot the structure of a problem by simplifying problems in a way which makes them look more like a previously seen problem., replacing \(u\) with \(e^x\).

\begin{equation} e^x = \sinh (x) \pm \sqrt {\sinh (x)^2 + 1} \end{equation}

The next step is to take the natural logarithm of both sides

\begin{equation} x = \ln \left ( \sinh (x) \pm \sqrt {\sinh (x)^2 + 1} \right ) \end{equation}

here, because \(a<\sqrt {a^2+1}\) 104104 This can be proved by considering the cases when \(a > 0\), \(a=0\) and \(a<0\) we cannot have the negative case (for the \(\pm \)), as the domain of the logarithm function requires that the input is greater than zero.

And we have found the inverse function!

\begin{equation} \operatorname {arsinh}(x) = \ln \left (x+\sqrt {x^2+1}\right ) \label {arsinhx} \end{equation}