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Question Number 20385 by tammi last updated on 26/Aug/17

$$\int \sqrt{16-9x^{2}dx}$$

Answered by ajfour last updated on 26/Aug/17

$$=3\int \sqrt{{\left(\frac{4}{3}\right)}^2-x^2}dx$$$$=\frac{3x}{2}\sqrt{{\left(\frac{4}{3}\right)}^2-x^2}+\frac{3\times 16}{2\times 9}{\sin }^{-1}\left(\frac{3x}{4}\right)-C$$$$=\frac{x}{2}\sqrt{16-9x^2}+\frac{8}{3}{\sin }^{-1}\left(\frac{3x}{4}\right)+C.$$

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