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$\require{cancel}$ I must find the length of the curve given by those 2 equations: $x = \cos^3t$ and $y = \sin^3t$ between $0$ and $2\pi$.

I can find their derivative: $\frac{dx}{dt} = -3\cos^2t\sin t$ and $\frac{dy}{dt} = 3\sin^2t\cos t$.

Then I can apply the formula (results on the right have been acquired with my calculator):

$$L_0^{2\pi} = \int_0^{2\pi}{\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}}dt = \int_0^{2\pi}{\sqrt{\left(-3\cos^2t\sin t\right)^2 + \left(3\sin^2t\cos t\right)}}dt = 6$$

$$L_0^{2\pi} = \int_0^{2\pi}{\sqrt{9\cos^4t\sin^2t + 9\sin^4t\cos^2t}}\,dt = 6$$

$$L_0^{2\pi} = 3\int_0^{2\pi}{\sqrt{\cos^2t\sin^2t\cdot(\cos^2t + \sin^2t)}}\,dt = 6$$

$$L_0^{2\pi} = 3\int_0^{2\pi}{\sqrt{\cos^2t\sin^2t}}\,dt = 6$$

$$L_0^{2\pi} = 3\int_0^{2\pi}{\cos t\sin t}\,dt = 0\,\,???$$

Let $z = \sin t \Rightarrow dz = \cos t\,dt$

For $t = 0, z = 0$ and for $t = 2\pi, z = 0$

$$L_0^{2\pi} = 3\int_0^0\cancel{\cos t}\cdot z\cdot\frac{1}{\cancel{\cos t}}\,dz = 3\int_0^0z\,dz = 0$$

$$L_0^{2\pi} = 3\left[\frac{z^2}{2}\right]_0^0 = 0$$

So how does the value of the integral changes only with a square root simplification (multiplying exponents $\frac{1}{2}$ and $2$)?

$\int_0^0$ doesn't make much sense to me but we weren't told how to deal with this kind of situation. How should I proceed to get the good answer?

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    $\begingroup$ The square root is a positive square root. That is, you are integrating not $\cos x \sin x$ but rather $|\cos x \sin x|$. This integral should give the desired result. $\endgroup$ Dec 4, 2017 at 0:06
  • $\begingroup$ One issue: You're missing the fact that, in general, $$\sqrt{\cos^2(t)\sin^2(t)} = |\cos(t)\sin(t)| \neq \cos(t)\sin(t)$$ $\endgroup$ Dec 4, 2017 at 0:07

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The problem is that $\sqrt{\cos^2 t\sin^2 t}$ is only $\cos t \sin t$ when $\cos t \sin t \ge 0$. Otherwise it's $-\cos t \sin t$.

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    $\begingroup$ Yes and $\cos t\sin t \ge 0$ when $t \in [0, \frac{\pi}{2}] \cup[\pi, \frac{3\pi}{2}]$ forcing me to split my integral resulting in no $\int_0^0$. Thanks ! $\endgroup$
    – Winter
    Dec 4, 2017 at 0:14

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