Show that $$\int _ { 0 } ^ { \pi } \mathrm { e } ^ { x } \sin x \mathrm {~d} x = \frac { 1 + \mathrm { e } ^ { \pi } } { 2 }$$ Given that $$I _ { n } = \int _ { 0 } ^ { \pi } \mathrm { e } ^ { x } \sin ^ { n } x \mathrm {~d} x$$ show that, for \(n \geqslant 2\), $$I _ { n } = n ( n - 1 ) \int _ { 0 } ^ { \pi } \mathrm { e } ^ { x } \cos ^ { 2 } x \sin ^ { n - 2 } x \mathrm {~d} x - n I _ { n }$$ and deduce that $$\left( n ^ { 2 } + 1 \right) I _ { n } = n ( n - 1 ) I _ { n - 2 } .$$ A curve has equation \(y = \mathrm { e } ^ { x } \sin ^ { 5 } x\). Find, in an exact form, the mean value of \(y\) over the interval \(0 \leqslant x \leqslant \pi\).