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Evaluate $\int \frac{x^{2}}{\sqrt{9-x^{2}}} d x$

Case 1: $\sqrt{a^{2}-x^{2}}$

$x=3 \sin \theta \quad-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$

$d x=3 \cos \theta d \theta$

$\sqrt{9-x^{2}}=3 \cos \theta$

$I= \int \frac{x^{2}}{\sqrt{2-x^{2}}} d x=\int \frac{(3 \sin \theta)^{2}}{3 \cos \theta} 3 \cos \theta d \theta$

$I=\int 9 \sin ^{2} \theta d \theta \quad \text { but } \sin ^{2} \theta=\frac{1}{2}[1-\cos (2 \theta)]$

$I=\int 9\left(\frac{1}{2}(1-\cos (2 \theta))\right) d \theta$

$=\frac{9}{2} \int(1-\cos 2 \theta) d \theta=\frac{9}{2}\left[\theta-\frac{1}{2} \sin (2\theta)\right]+c$

$* \sin (2 \theta)=2 \sin \theta \cos \theta$

$I=\frac{9}{2}\left[\theta-\frac{1}{2}[2 \sin \theta \cos \theta]\right] + c$

$=\frac{9}{2}[\theta-\sin \theta \cos \theta]+c$

$\sin \theta=\frac{x}{3}$

$I=\int \frac{x^{2}}{\sqrt{9-x^{2}}} d x$

$I=\int \frac{x^{2}}{\sqrt{9-x^{2}}} d x=\frac{9}{2}[\theta-\sin \cos \theta]+c$

$\theta=\sin ^{-1}\left(\frac{x}{3}\right), \cos \theta=\frac{\sqrt{9-x^{2}}}{3}$

$=\frac{9}{2}\left[\sin ^{-1}\left((\frac{x}{3})-\frac{x}{3} \frac{\sqrt{9-{x}^{2}}}{3}\right)\right]$

$=\frac{9}{2}\left[\sin ^{-1}\left(\frac{x}{3}\right)-\frac{x \sqrt{9-{x}^{2}}}{9}\right]$