7 It is given that, for non-negative integers \(n , I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { n } x \mathrm {~d} x\).
- Show that \(I _ { n } = \frac { n - 1 } { n } I _ { n - 2 }\) for \(n \geqslant 2\).
- Explain why \(I _ { 2 n + 1 } < I _ { 2 n - 1 }\).
- It is given that \(I _ { 2 n + 1 } < I _ { 2 n } < I _ { 2 n - 1 }\). Take \(n = 5\) to find an interval within which the value of \(\pi\) lies.