Question

To determine that \({\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right)\).

Short answer

The prove has been given below.

Step-by-step solution

Given data

The equation to be proved is \({\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right)\).

Identity of hyperbolic function

Hyperbolic function:

\({\cosh ^2}y - {\sinh ^2}y = 1\)

\(\cosh y + \sinh y = {e^y}\)

Prove of the equation \({\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right)\)

To prove the above equation let\(y = {\sinh ^{ - 1}}x\).

\( \Rightarrow x = \sinh y\)

Now, substitute \(x = \sinh y\) in \({\cosh ^2}y - {\sinh ^2}y = 1\).

\(\begin{array}{c}{\cosh ^2}y - {x^2} = 1\\{\cosh ^2}y = 1 + {x^2}\\\cosh y = \sqrt {1 + {x^2}} \end{array}\)

Now, substitute \(\cosh y = \sqrt {1 + {x^2}} \) and \(\sinh y = x\) in \(\cosh y + \sinh y = {e^y}\).

\(\begin{array}{c}\sqrt {1 + {x^2}} + x = {e^y}\\{e^y} = x + \sqrt {1 + {x^2}} \end{array}\)

Take natural logarithm on both sides.

\(\begin{array}{c}y = \ln \left( {x + \sqrt {{x^2} + 1} } \right)\\{\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right)\\y = {\sinh ^{ - 1}}x\end{array}\)

Hence, the result is proved.