8.06d

Use the definition $\cosh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)$ to show that $\cosh 2 x = 2 \cosh ^ { 2 } x - 1$.
(2 marks)
1. The arc of the curve $y = \cosh x$ between $x = 0$ and $x = \ln a$ is rotated through $2 \pi$ radians about the $x$-axis. Show that $S$, the surface area generated, is given by $$S = 2 \pi \int _ { 0 } ^ { \ln a } \cosh ^ { 2 } x \mathrm {~d} x$$
2. Hence show that $$S = \pi \left( \ln a + \frac { a ^ { 4 } - 1 } { 4 a ^ { 2 } } \right)$$