Respuesta :

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[tex]\displaystyle\frac1{\pi-(-\pi)}\int_{-\pi}^\pi9\sin^2x\cos^3x\,\mathrm dx[/tex]
[tex]=\displaystyle\frac9{2\pi}\int_{-\pi}^\pi\sin^2x(1-\sin^2x)\cos x\,\mathrm dx[/tex]
[tex]=\displaystyle\frac9{2\pi}\int_0^0 y^2(1-y^2)\,\mathrm dy[/tex]

where we let [tex]y=\sin x[/tex]. The remaining integral reduces to 0.

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