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**o_O** Note that $\displaystyle s = \int v dt$ where s represents distance. So:

$\displaystyle s = \int \frac{1}{\sqrt{9 - t^{2}}}dt$

Do a substitution of:

$\displaystyle t = 3\sin \theta$

$\displaystyle dt = 3\cos \theta d\theta$

So you get:

$\displaystyle s = \int \frac{3\cos \theta d\theta}{\sqrt{9 - 9\sin^{2} \theta}} = \int \frac{3\cos \theta d\theta}{3 \sqrt{1 - \sin^{2} \theta}}$

Can you see where you can go with the $\displaystyle 1 - \sin^{2} \theta$?

Note that there will be the constant after integrating. They give you the point (0, 10) to allow you to solve for it.