Show that
$$\frac { 1 } { 4 } ( \cosh 4 x + 2 \cosh 2 x + 1 ) = \cosh ^ { 2 } x \cosh 2 x$$
Show that, if \(y = \cosh ^ { 2 } x\), then
$$1 + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = \cosh ^ { 2 } 2 x$$
The arc of the curve \(y = \cosh ^ { 2 } x\) between the points where \(x = 0\) and \(x = \ln 2\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Show that the area \(S\) of the curved surface formed is given by
$$S = \frac { \pi } { 256 } ( a \ln 2 + b )$$
where \(a\) and \(b\) are integers.