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Let $x = 2\cos{2\theta}$. Then, $dx = -4\sin{2\theta} d\theta$. We also need to change the limits of integration.
When $x = -2$, $2\cos{2\theta} = -2$, so $\cos{2\theta} = -1$, which means $2\theta = \pi$, and $\theta = \frac{\pi}{2}$. When $x = 2$, $2\cos{2\theta} = 2$, so $\cos{2\theta} = 1$, which means $2\theta = 0$, and $\theta = 0$.
The integral becomes: $$ \int_{\pi/2}^{0} \sqrt{\frac{2-2\cos{2\theta}}{2+2\cos{2\theta}}} (-4\sin{2\theta}) d\theta = \int_{\pi/2}^{0} \sqrt{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}} (-4\sin{2\theta}) d\theta $$
Using the identities $1 - \cos{2\theta} = 2\sin^2{\theta}$ and $1 + \cos{2\theta} = 2\cos^2{\theta}$, we have: $$ \int_{\pi/2}^{0} \sqrt{\frac{2\sin^2{\theta}}{2\cos^2{\theta}}} (-4\sin{2\theta}) d\theta = \int_{\pi/2}^{0} \sqrt{\tan^2{\theta}} (-4\sin{2\theta}) d\theta = \int_{\pi/2}^{0} \tan{\theta} (-4\sin{2\theta}) d\theta $$
Since $\sin{2\theta} = 2\sin{\theta}\cos{\theta}$, we have: $$ \int_{\pi/2}^{0} \frac{\sin{\theta}}{\cos{\theta}} (-4(2\sin{\theta}\cos{\theta})) d\theta = \int_{\pi/2}^{0} \frac{\sin{\theta}}{\cos{\theta}} (-8\sin{\theta}\cos{\theta}) d\theta = \int_{\pi/2}^{0} -8\sin^2{\theta} d\theta $$
Changing the limits of integration, we get: $$ 8\int_{0}^{\pi/2} \sin^2{\theta} d\theta $$ Using the identity $\sin^2{\theta} = \frac{1 - \cos{2\theta}}{2}$, we have: $$ 8\int_{0}^{\pi/2} \frac{1 - \cos{2\theta}}{2} d\theta = 4\int_{0}^{\pi/2} (1 - \cos{2\theta}) d\theta $$
$$ 4\int_{0}^{\pi/2} (1 - \cos{2\theta}) d\theta = 4\left[\theta - \frac{\sin{2\theta}}{2}\right]_{0}^{\pi/2} = 4\left[\left(\frac{\pi}{2} - \frac{\sin{\pi}}{2}\right) - \left(0 - \frac{\sin{0}}{2}\right)\right] = 4\left[\frac{\pi}{2} - 0 - 0 + 0\right] = 4\left(\frac{\pi}{2}\right) = 2\pi $$
Final Answer: $2\pi$
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