A calculus of the absurd

15.2.3 Inverse function of \(\sinh (x)\)
  • Example 15.2.1 Finding the inverse function of \(\sinh (x)\).

From the definition of \(\sinh (x)\) we know that

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

The right-hand side is actually a quadratic in \(e^x\), which becomes more readily apparent114114 In general, if we have an expression containing \(x^y\), \(x^{-y}\) and no other powers of \(x\), it’s worth considering whether there is a hidden quadratic to be found. if we multiply 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 substitution115115 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 \(a<\sqrt {a^2+1}\) 116116 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}