1. ## Integration: Length

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

2. You may want to do that all again being careful about what variable your integrating with respect to and how exactly you take the square root of a sum

3. A little pointer.

$\displaystyle \cos^2 2t + 8 \cos^2 t$
$\displaystyle ( 2 \cos^2 t - 1 )^2 + 8 \cos^2 t$
$\displaystyle 4 \cos^4 t - 4 \cos^2 t + 1 + 8 \cos^2 t$
$\displaystyle 4 \cos^4 t + 4 \cos^2 t + 1$
$\displaystyle ( 2 \cos^2 t + 1)^2$

the rest is easy.

4. Originally Posted by bobak
You may want to do that all again being careful about what variable your integrating with respect to and how exactly you take the square root of a sum
Sorry, it is with respect to $\displaystyle t$. I copied my latex code from my other post so forgot to change the letter. It's edited now.