# 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

$$\sinh (x) = \frac {e^x-e^{-x}}{2}$$

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 $$\frac {e^{x}}{e^{x}}$$.

$$\sinh (x) = \frac {e^{2x}-1}{2e^{x}}$$

Here we can substitute $$u=e^{x}$$

$$\sinh (x) = \frac {u^2-1}{2u}$$

And then we can rearrange this to give a quadratic in terms of $$u$$.

$$u^2 - 2u\sinh (x) - 1 = 0$$

By applying the quadratic formula, we can solve this for $$u$$

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

We can then simplify this a bit

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

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 $$e^x$$.

$$e^x = \sinh (x) \pm \sqrt {\sinh (x)^2 + 1}$$

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

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

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!

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