Question

∫ (1-sin²3t)cos3t du

Answer 1

Substitute
`y=\sin3t`, so that
`\mathrm dy=3\cos3t\,\mathrm dt` (not
`\mathrm du`!). Then

`\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`