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

Case 1: \(\sqrt{a^{2}-x^{2}}\)

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

\(d x=3 \cos \theta d \theta\)

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

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

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

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

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

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

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

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

\(\sin \theta=\frac{x}{3}\)

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

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

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

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

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

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