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 $C_{k}$.
I thought this would be fairly straightforward but im a little confused. Do I start out by doing a trig substitution?
On
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
$$\int_{-2}^{2} x^{2k} \ \sqrt{4-x^2} \ dx \ = \ 4^{k+1} \cdot \frac{ [2k-1] \cdot \ldots \cdot 3 \cdot 1}{2^{k+1} \cdot \ (k+1)!} \cdot \pi $$
$$\Rightarrow \ C_k \ = \ \frac{4^{k+1} \cdot \frac{ [2k-1] \ \cdot \ [2k-3] \ \cdot \ \ldots \ \cdot \ 3 \ \cdot \ 1}{2^{k+1} \cdot \ (k+1)!} \cdot \pi}{4^k \cdot \frac{ [2k-3] \ \cdot \ \ldots \ \cdot \ 3 \ \cdot 1}{2^k \ \cdot \ k!} \cdot \pi} \ = \ \frac{4 \cdot (2k-1)}{2 \cdot (k+1)} \ = \ \frac{2 (2k-1)}{k+1} \ . $$
On
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 $C_k= \dfrac{2(2 k-1)}{k+1} .$
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
$$C_k = \displaystyle \frac{4k-2}{k+1}.$$