Calculus Reduction Formula 
For any integer $k > 0$, show the reduction formula
  $$\int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx = C_k \int^{2}_{-2} x^{2k-2} \sqrt{4-x^2} \, dx$$
  for some constant $C_{k}$.

(original image)
I thought this would be fairly straightforward but im a little confused. Do I start out by doing a trig substitution?

Let $I_k = \displaystyle \int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx$.
Let $u = x^{2k-1}$ and $dv = x\sqrt{4-x^2} \, dx$.
We then have $du = (2k-1)x^{2k-2} \, dx$ and 
$\begin{align*}
v &= \int x\sqrt{4-x^2} \, dx\\
&=\displaystyle -\frac{1}{3} (4-x^2)^{3/2}\\
&=-\frac{1}{3}(4-x^2)\sqrt{4-x^2}\end{align*}$
Applying integration by parts $\left (\int u \, dv = uv - \int v \, du \right )$:
$$\begin{align*}
I_k&= \left [ x^{2k-1} \times -\frac{1}{3}\overbrace{(4-x^2)^{3/2}}^{\text{this term becomes 0}} \right ]^{2}_{-2} + \frac{1}{3}(2k-1)\int^{2}_{-2} x^{2k-2}(4-x^2)\sqrt{4-x^2} \, dx\\
&= 0 + \frac{1}{3}(2k-1)\int^{2}_{-2}4x^{2k-2}\sqrt{4-x^2} - x^{2k}\sqrt{4-x^2} \, dx\\
&= \frac{4}{3}(2k-1)\overbrace{\int^{2}_{-2}4x^{2k-2}\sqrt{4-x^2} \, dx}^{I_{k-1}} - \frac{1}{3}(2k-1) \overbrace{\int^{2}_{-2} - x^{2k}\sqrt{4-x^2} \, dx}^{I_k}\\[10pt]
\therefore 3I_k &= 4(2k-1)I_{k-1} - (2k-1)I_{k}\\\\
\end{align*}$$
Rearranging yields $I_k = \displaystyle \frac{4k-2}{k+1}I_{k-1}$, so 
$$C_k = \displaystyle \frac{4k-2}{k+1}.$$
