Question

Give an alternative solution to Example 3 by letting \(y = {\bf{sin}}{{\bf{h}}^{ - {\bf{1}}}}x\) and then using Exercise 13 and Example 1(a) with x replaced by y.

Short answer

The value of \({\sinh ^{ - 1}}x\) is \(\ln \left( {x + \sqrt {1 + {x^2}} } \right)\).

Step-by-step solution

Step 1:Simplify the problem using the trigonometric identity

Let \(y = {\sinh ^{ - 1}}x\), then\(x = \sinh y\).

Using the hyperbolic relation,

\(\begin{aligned}{\cosh ^2}y - {\sinh ^2}y &= 1\\{\cosh ^2}y &= 1 + {\sinh ^2}y\\\cosh y &= \sqrt {1 + {{\sinh }^2}y} \\ &= \sqrt {1 + {x^2}} \end{aligned}\)

Use the identity given in Example 13

The given relation in example 13 is \({e^y} = \sinh y + \cosh y\).

Substitute \(x\) for \(\sinh y\) and \(\sqrt {1 + {x^2}} \) for \(\cosh y\).

\(\begin{aligned} {{e}^{y}} &=x+\sqrt{1+{{x}^{2}}} \\ y &=\ln \left( x+\sqrt{1+{{x}^{2}}} \right) \end{aligned}\)

So, the value of \({\sinh ^{ - 1}}x\) is \(\ln \left( {x + \sqrt {1 + {x^2}} } \right)\).