Prove, from definitions involving exponentials, that
$$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
Given that \(\sinh x = \tan y\), where \(- \frac { 1 } { 2 } \pi < y < \frac { 1 } { 2 } \pi\), show that
(A) \(\tanh x = \sin y\),
(B) \(x = \ln ( \tan y + \sec y )\).
Given that \(y = \operatorname { artanh } x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).
Hence show that \(\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { 1 - x ^ { 2 } } \mathrm {~d} x = 2 \operatorname { artanh } \frac { 1 } { 2 }\).
Express \(\frac { 1 } { 1 - x ^ { 2 } }\) in partial fractions and hence find an expression for \(\int \frac { 1 } { 1 - x ^ { 2 } } \mathrm {~d} x\) in terms of logarithms.
Use the results in parts (i) and (ii) to show that \(\operatorname { artanh } \frac { 1 } { 2 } = \frac { 1 } { 2 } \ln 3\).