All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
\int \frac{1}{\sqrt{9-x^{2}}} dx
= sin^{-1}(\frac{x}{3})+C

(ii)
\frac{1}{\sqrt{9-x^{2}}}

= \frac{1}{\sqrt{9} \sqrt{1-\frac{x^{2}}{9}}}

= \frac{1}{3 \sqrt{1-\frac{x^{2}}{9}}}

= \frac{1}{3}[1-\frac{1}{2}(-\frac{x^{2}}{9})+\frac{-\frac{1}{2} (-\frac{3}{2})}{2!}(-\frac{x^{2}}{9})^{2}+\frac{-\frac{1}{2} (-\frac{3}{2})(-\frac{5}{2})}{3!}(-\frac{x^{2}}{9})^{3}+...

= \frac{1}{3}[1+\frac{x^{2}}{18}+\frac{3x^{4}}{648}+\frac{5x^{6}}{11664}+...]

= \frac{1}{3}+\frac{x^{2}}{54}+\frac{x^{4}}{648}+\frac{5x^{6}}{34992}+...

(iii)
Note that sin^{-1}(\frac{x}{3}) = \int \frac{1}{\sqrt{9-x^{2}}} dx
\int \frac{1}{\sqrt{9-x^{2}}} dx

\approx \int \frac{1}{3}+\frac{x^{2}}{54}+\frac{x^{4}}{648}+\frac{5x^{6}}{34992} dx

=\frac{x}{3}+\frac{x^{3}}{162}+\frac{x^{5}}{3240}+\frac{5x^{7}}{244944} +C

When x=0, C=0

Therefore, sin^{-1}(\frac{x}{3}) \approx \frac{x}{3}+\frac{x^{3}}{162}+\frac{x^{5}}{3240}+\frac{5x^{7}}{244944}

Personal Comments:
Some students still forgot to add the arbitrary constants. 🙁
Students should know how to utilise the MF15 formula to do the integration. The rest of the question is quite standard, though the calculations are ready tedious and hard here.

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