2 It is given that \(\alpha\) is the only real root of the equation \(x ^ { 5 } + 2 x - 28 = 0\) and that \(1.8 < \alpha < 2\).
- The iteration \(x _ { n + 1 } = \sqrt [ 5 ] { 28 - 2 x _ { n } }\), with \(x _ { 1 } = 1.9\), is to be used to find \(\alpha\). Find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving the answers correct to 7 decimal places.
- The error \(e _ { n }\) is defined by \(e _ { n } = \alpha - x _ { n }\). Given that \(\alpha = 1.8915749\), correct to 7 decimal places, evaluate \(\frac { e _ { 3 } } { e _ { 2 } }\) and \(\frac { e _ { 4 } } { e _ { 3 } }\). Comment on these values in relation to the gradient of the curve with equation \(y = \sqrt [ 5 ] { 28 - 2 x }\) at \(x = \alpha\).
- Prove that the derivative of \(\sin ^ { - 1 } x\) is \(\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
- Given that
$$\sin ^ { - 1 } 2 x + \sin ^ { - 1 } y = \frac { 1 } { 2 } \pi$$
find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = \frac { 1 } { 4 }\).
- By means of a suitable substitution, show that
$$\int \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x$$
can be transformed to \(\int \cosh ^ { 2 } \theta \mathrm {~d} \theta\).
- Hence show that \(\int \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x = \frac { 1 } { 2 } x \sqrt { x ^ { 2 } - 1 } + \frac { 1 } { 2 } \cosh ^ { - 1 } x + c\).