Let \[ I = \int {\frac{{dx}}{{\sqrt {16 - 9{x^2}} }}} = \int {\frac{{dx}}{{\sqrt {{{(4)}^2} - {{(3x)}^2}} }}} dx = \frac{1}{3}{\sin ^{ - 1}}\left( {\frac{{3x}}{4}} \right) + C\]

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