Show that
$$\frac { \mathrm { d } } { \mathrm {~d} u } \left( 2 u \sqrt { 1 + 4 u ^ { 2 } } + \sinh ^ { - 1 } 2 u \right) = k \sqrt { 1 + 4 u ^ { 2 } }$$
where \(k\) is an integer.
Hence show that
$$\int _ { 0 } ^ { 1 } \sqrt { 1 + 4 u ^ { 2 } } \mathrm {~d} u = p \sqrt { 5 } + q \sinh ^ { - 1 } 2$$
where \(p\) and \(q\) are rational numbers.
The arc of the curve with equation \(y = \frac { 1 } { 2 } \cos 4 x\) between the points where \(x = 0\) and \(x = \frac { \pi } { 8 }\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
Show that the area \(S\) of the curved surface formed is given by
$$S = \pi \int _ { 0 } ^ { \frac { \pi } { 8 } } \cos 4 x \sqrt { 1 + 4 \sin ^ { 2 } 4 x } \mathrm {~d} x$$
Use the substitution \(u = \sin 4 x\) to find the exact value of \(S\).