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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