Question

Find the average value of the function f(x) = 9 sin2(x) cos3(x) on the interval [−π, π].

Answer 1

`\displaystyle\frac1{\pi-(-\pi)}\int_(-\pi)^\pi9\sin^2x\cos^3x\,\mathrm dx`

`=\displaystyle\frac9{2\pi}\int_(-\pi)^\pi\sin^2x(1-\sin^2x)\cos x\,\mathrm dx`

`=\displaystyle\frac9{2\pi}\int_0^0 y^2(1-y^2)\,\mathrm dy`

where we let
`y=\sin x`. The remaining integral reduces to 0.