<s>[INST] 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 $C_{k}$.

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I thought this would be fairly straightforward but im a little confused. Do I start out by doing a trig substitution?
 [/INST] I like to find the reduction formula for  the integral by rationalization followed by integration by parts.
$$
\begin{aligned}
I_{k} &=\int_{-2}^{2} x^{2 k} \sqrt{4-x^{2}} d x \\
&=\int_{-2}^{2} \frac{x^{2 k}\left(4-x^{2}\right)}{\sqrt{4-x^{2}}} d x \\
&=-\int_{-2}^{2} x^{2 k-1}\left(4-x^{2}\right) d\left(\sqrt{4-x^{2}}\right) \\
&=-\left[x^{2 k-1}\left(4-x^{2}\right)^{\frac{3}{2}}\right]_{-2}^{2}+\int_{-2}^{2}\left[4(2 k-1) x^{2 k-2}-(2 k+1) x^{2 k} \right] \sqrt{4-x^2}dx\\
&=4(2 k-1) I_{k-1}-(2 k+1) I_{k} \\\\
I_{k} &=\frac{2(2 k-1)}{k+1} I_{k-1}
\end{aligned} \tag*{} 
\\$$
Hence $C_k= \dfrac{2(2 k-1)}{k+1} .$
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