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Bởi vì đường cong là đối xứng nên ta có
$L = 4\int\limits_{0}^{\pi/2}\sqrt{(x')^2+(y')^2}dt$
$L = 4\int\limits_{0}^{\pi/2}\sqrt{(3a\cos^2t\sin t)^2+(3a\sin^2t\cos t)^2}dt$
$L = 12a\int\limits_{0}^{\pi/2}\sqrt{\cos^4t\sin^2 t+\sin^4t\cos^2t}dt$
$L = 12a\int\limits_{0}^{\pi/2}\sqrt{\sin^2t\cos^2t}dt$
$L = 12a\int\limits_{0}^{\pi/2}\sin t\cos tdt$
$L = 6a\int\limits_{0}^{\pi/2}\sin 2tdt$
$L = 6a\left[ {-1/2\cos2t} \right]_{0}^{\pi/2}$
$L = 6a$
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