Substitute [tex]y=\sin3t[/tex], so that [tex]\mathrm dy=3\cos3t\,\mathrm dt[/tex] (not [tex]\mathrm du[/tex]!). Then
[tex]\displaystyle\int(1-\sin^23t)\cos3t\,\mathrm dt=\frac13\int(1-y^2)\,\mathrm dy=\frac y3-\frac{y^3}9+C=\frac{\sin3t}3-\frac{\sin^3t}9+C[/tex]
