## Calculus 8th Edition

$$\displaystyle{\int\frac{x^2}{\sqrt{9-x^2}}dx=\frac{9\arcsin\left(\frac{x}{3}\right)-x\sqrt{9-x^2}}{2}+C}\\$$
$\displaystyle{I=\int\frac{x^2}{\sqrt{9-x^2}}dx}\\$ $\displaystyle \left[\begin{array}{ll} x=3\sin\theta & x^2=9\sin^2\theta \\ & \\ \frac{dx}{d\theta}=3\cos\theta & dx=3\cos\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int\frac{9\sin^2\theta}{\sqrt{9-9\sin^2\theta}}3\cos\theta\ d\theta}\\ \displaystyle{I=\int\frac{9\sin^2\theta}{\sqrt{9\left(1-\sin^2\theta\right)}}3\cos\theta\ d\theta}\\ \displaystyle{I=\int9\sin^2\theta}\\ \displaystyle{I=\frac{9}{2}\int1-\cos{2\theta}\ d\theta}\\ \displaystyle{I=\frac{9}{2}\left(\theta-\frac{1}{2}\sin{2\theta}\right)+C}\\ \displaystyle{I=\frac{9}{2}\theta-\frac{9}{4}\sin{2\theta}+C}\\ \displaystyle{I=\frac{9}{2}\theta-\frac{9}{2}\sin{\theta}\cos\theta+C}$ $\theta=\arcsin\left(\frac{x}{3}\right)\\ \sin\theta=\frac{x}{3}\\ \cos\theta=\frac{\sqrt{9-x^2}}{3}\\$ $\displaystyle{I=\frac{9}{2}\arcsin\left(\frac{x}{3}\right)-\frac{9}{2}\times\frac{x}{3}\times\frac{\sqrt{9-x^2}}{3}+C}\\ \displaystyle{I=\frac{9\arcsin\left(\frac{x}{3}\right)-x\sqrt{9-x^2}}{2}+C}$