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(i)

(ii)
Since $x=asin\theta$, we have $dx=acos\theta ~d\theta$
When $x= \frac{\sqrt{3}a}{2}, \theta = \frac{\pi}{3}$
When $x= \frac{a}{2}, \theta = \frac{\pi}{6}$

$\int_{\pi / 6}^{\pi / 3} \sqrt{1-sin^{2}\theta}~(acos\theta) ~d\theta$

$= a \int_{\pi / 6}^{\pi / 3} cos^{2} \theta ~d \theta$

$= a \int_{\pi / 6}^{\pi / 3} \frac{1+cos2\theta}{2} ~d \theta$

$= \frac{a}{2}[\theta + \frac{sin 2 \theta}{2}]\biggl|_{\pi / 6}^{\pi / 3}$

$= \frac{a \pi}{12}$

### KS Comments:

The function is a periodic curve and some students do have problem reading such functions. They should be alert of the domain that they are required to draw too. Some students were unsure how the hint that $f(x+3a)=f(x)$ should help. The parametric integration should be quite simple to handle as students just need to be cautious and introduce the double angle formula from MF15 effectively.

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