Problem

(a) Starting from the definition of cosh in terms of exponentials, prove that

$$2\cosh^2A=\cosh 2A+1.$$

The curve $C$ has parametric equations

$$x=2\cosh 2t+3t,\qquad y=\frac{3}{2}\cosh 2t-4t,\qquad -\frac{1}{2}\le t\le\frac{1}{2}.$$

The area of the surface generated when $C$ is rotated through $2\pi$ radians about the $y$-axis is denoted by $A$.

(b) (i) Show that

$$A=10\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\big(2\cosh 2t+3t\big)\cosh 2t\,\mathrm{d}t.$$

(ii) Hence find $A$ in terms of $\pi$ and $e$.