6.
$$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 2 } } x ^ { n } \sin x \mathrm {~d} x$$
- Show that for \(n \geq 2\)
$$I _ { n } = n \left( \frac { \pi } { 2 } \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 }$$
- Hence obtain \(I _ { 3 }\), giving your answers in terms of \(\pi\).