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Question Number 40952 by Raj Singh last updated on 30/Jul/18

Commented by abdo mathsup 649 cc last updated on 30/Jul/18

$$letI=\int \frac{dx}{{(9-x^2)}^{\frac{3}{2}}}$$$$I{=}_{x=3sin\theta }\int \frac{3cos\theta d\theta }{9^{\frac{3}{2}}{(1-{sin}^2\theta )}^{\frac{3}{2}}}$$$$=\frac{1}{9}\int \frac{cos\theta }{{cos}^3\theta }d\theta =\frac{1}{9}\int \frac{d\theta }{{cos}^2\theta }$$$$9I=\int (1+{tan}^2\theta )d\theta {=}_{tan\theta =u}\int (1+u^2)\frac{du}{1+u^2}$$$$=u+c=tan\theta +c=tan\left(arcsin\left(\frac{x}{3}\right)\right)+c$$$$I=\frac{1}{9}tan\left(arcsin\left(\frac{x}{3}\right)\right)+c.$$

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