8.
$$y = \arccos ( 2 \sqrt { x } )$$
- Determine \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
- Show that
$$\int y \mathrm {~d} x = x \arccos ( 2 \sqrt { x } ) + \int \frac { \sqrt { x } } { \sqrt { 1 - 4 x } } \mathrm {~d} x$$
- Use the substitution \(\sqrt { x } = \frac { 1 } { 2 } \cos \theta\) to show that
$$\int _ { 0 } ^ { \frac { 1 } { 8 } } \frac { \sqrt { x } } { \sqrt { 1 - 4 x } } \mathrm {~d} x = \frac { 1 } { 4 } \int _ { a } ^ { b } \cos ^ { 2 } \theta \mathrm {~d} \theta$$
where \(a\) and \(b\) are limits to be determined.
- Hence, determine the exact value of
$$\int _ { 0 } ^ { \frac { 1 } { 8 } } \arccos ( 2 \sqrt { x } ) d x$$