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$I=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(x){\rm d}x$$J=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(-x){\rm d}x$Ta có \begin{align*}I+J&=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}\sqrt{2-2\cos 2x}{\rm d}x\\ &=2\int_{\frac{-3\pi}2}^{\frac{3\pi}2}\sqrt{\sin^2x}{\rm d}x\\ &= 4\int_0^{\frac{3\pi}2}\left|\sin x\right|{\rm d}x \\ &=4\int_0^{\pi}\left|\sin x\right|{\rm d}x+4\int_{\pi}^{\frac{3\pi}2}\left|\sin x\right|{\rm d}x \\ &=4\int_0^{\pi}\sin x{\rm d}x+4\int_{\pi}^{\frac{3\pi}2}(-\sin x){\rm d}x \\ &=-4\cos x\bigg|_0^{\pi}+4\cos x\bigg|_{\pi}^{\frac{3\pi}2}\\&=24\end{align*}Mặt khác, khi đặt $t=-x$ thì ta có:$I=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(x){\rm d}x=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(t){\rm d}t=\int_{\frac{3\pi}2}^{\frac{-3\pi}2}f(-x){\rm d}\left(-x\right)=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(-x){\rm d}x=J$Suy ra $I=J=12$, vậy tích phân cần tìm là 12

$I=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(x){\rm d}x$$J=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(-x){\rm d}x$Ta có \begin{align*}I+J&=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}\sqrt{2-2\cos 2x}{\rm d}x\\ &=2\int_{\frac{-3\pi}2}^{\frac{3\pi}2}\sqrt{\sin^2x}{\rm d}x\\ &= 4\int_0^{\frac{3\pi}2}\left|\sin x\right|{\rm d}x \\ &=4\int_0^{\pi}\left|\sin x\right|{\rm d}x+4\int_{\pi}^{\frac{3\pi}2}\left|\sin x\right|{\rm d}x \\ &=4\int_0^{\pi}\sin x{\rm d}x+4\int_{\pi}^{\frac{3\pi}2}(-\sin x){\rm d}x \\ &=-4\cos x\bigg|_0^{\pi}+4\cos x\bigg|_{\pi}^{\frac{3\pi}2}\\&=12\end{align*}Mặt khác, khi đặt $t=-x$ thì ta có:$I=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(x){\rm d}x=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(t){\rm d}t=\int_{\frac{3\pi}2}^{\frac{-3\pi}2}f(-x){\rm d}\left(-x\right)=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(-x){\rm d}x=J$Suy ra $I=J=6$, vậy tích phân cần tìm là 6