In case you're wondering how to do the first one without trig sub, let $\displaystyle u=16-4x^2\implies x^2=\tfrac{1}{4}(16-u)$. Therefore, $\displaystyle du=-8x\,dx \implies -\frac{1}{8}\,du=dx$.
Therefore,
$\displaystyle \displaystyle\int x^3\sqrt{16-4x^2}\,dx \xrightarrow{u=16-x^2}{}-\tfrac{1}{8}\int x^2\sqrt{u}\,du=-\tfrac{1}{32}\int (16-u)\sqrt{u}\,du$
which can be easily integrated.