Q: $\displaystyle x=\sin t \cos t$, $\displaystyle 2\sqrt 2 \sin t$

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My method:

$\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t} = \cos ^2 t -\sin ^2 t = \cos t$

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t} = 2 \sqrt 2 \cos t $

$\displaystyle \therefore \displaystyle\int^\frac{\pi}{3}_0 ((\cos 2t)^2 + (2 \sqrt 2 \cos t)^2)^\frac{1}{2} \, \mathrm{d}t$

$\displaystyle \therefore \displaystyle\int^\frac{\pi}{3}_0 ((\cos 2t) + (2 \sqrt 2 \cos t)) \, \mathrm{d}t$

$\displaystyle = 2 \sin 2t + 2 \sqrt 2 \sin t$

Inserting limits does not give the correct answer. Where have I gone wrong?

If possible, can someone redo method and see if they get the same.

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Correct Answer: $\displaystyle \frac{1}{4} \sqrt 3 + \frac{2}{3} \pi$