<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] It took a little time to check through this, but here is an approach using the substitution I'd proposed in the comments.  Applying $\ \sin\theta = \frac{x}{2}$ , we have
$$\int_{-2}^{2} x^{2k} \ \sqrt{4-x^2} \ dx \ \rightarrow \ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2^{2k} \cdot \sin^{2k}\theta \cdot (2 \cos\theta) \cdot (2 \cos\theta \ d\theta )$$
$$= \ 4^{k+1}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}  \sin^{2k}\theta \   \cos^2\theta \ d\theta \ = \ 4^{k+1}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}  \sin^{2k} \theta \ - \ \sin^{2k+2} \theta \ d\theta \ ,$$
having applied the Pythagorean Identity in this last stage.  I will use the result (which I won't derive here) 
$$ \ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}  \sin^{2k}\theta \ d\theta \ = \ \frac{\sqrt{\pi} \cdot \Gamma(k+\frac{1}{2})}{\Gamma(k+1)} \ = \ \frac{\sqrt{\pi} \cdot (\frac{2k-1}{2} \cdot \ldots \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi})}{k!} $$
$$= \ \frac{  ([2k-1] \cdot \ldots \cdot 3 \cdot 1)}{2^k \cdot k!} \cdot \pi $$
$$\Rightarrow \ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}  \sin^{2k} \theta \ - \ \sin^{2k+2} \theta \ d\theta \ =  \ [ \frac{  ([2k-1] \cdot \ldots \cdot 3 \cdot 1)}{2^k \cdot k!}  -  \frac{  ([2k+1] \cdot \ldots \cdot 3 \cdot 1)}{2^{k+1} \cdot (k+1)!} ] \cdot \pi$$
$$=  \  \frac{[  \ 2(k+1) \ - \ (2k+1) \ ] \cdot ([2k-1] \cdot \ldots \cdot 3 \cdot 1)}{2^{k+1} \cdot (k+1)!}  \cdot \pi \ = \ \frac{ [2k-1] \cdot \ldots \cdot 3 \cdot 1}{2^{k+1} \cdot (k+1)!}  \cdot \pi \ .$$
Our original integral is then  
$$\int_{-2}^{2} x^{2k} \ \sqrt{4-x^2} \ dx \ = \ 4^{k+1}  \cdot  \frac{ [2k-1] \cdot \ldots \cdot 3 \cdot 1}{2^{k+1}  \cdot  \ (k+1)!}  \cdot \pi   $$
$$\Rightarrow \ C_k \ = \  \frac{4^{k+1}  \cdot  \frac{ [2k-1] \ \cdot \ [2k-3] \ \cdot \ \ldots \ \cdot \ 3  \ \cdot \ 1}{2^{k+1} \cdot \ (k+1)!}  \cdot \pi}{4^k  \cdot  \frac{ [2k-3] \ \cdot \  \ldots \ \cdot  \ 3 \ \cdot 1}{2^k \ \cdot \ k!}  \cdot \pi} \ = \ \frac{4 \cdot (2k-1)}{2 \cdot (k+1)} \ = \ \frac{2  (2k-1)}{k+1} \ . $$
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