## Calculus 8th Edition

$$\displaystyle{\int_{0}^{0.6}\frac{x^2}{\sqrt{9-25x^2}}dx=\frac{9\pi}{500}}\\$$
$\displaystyle{I=\int_{0}^{0.6}\frac{x^2}{\sqrt{9-25x^2}}dx}\\$ $\displaystyle \left[\begin{array}{ll} 5x=3\sin\theta & 25x^2=9\sin^2\theta \\ & \\ \frac{dx}{d\theta}=\frac{3}{5}\cos\theta & dx=\frac{3}{5}\cos\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int_{0}^{\frac{\pi}{2}}\frac{9\sin^2\theta}{25\sqrt{9-9\sin^2\theta}}\times\frac35\cos\theta\ d\theta}\\ \displaystyle{I=\int_{0}^{\frac{\pi}{2}}\frac{9\sin^2\theta}{25\sqrt{9\left(1-\sin^2\theta\right)}}\times\frac35\cos\theta\ d\theta}\\ \displaystyle{I=\int_{0}^{\frac{\pi}{2}}\frac{9}{125}\sin^2\theta}\\ \displaystyle{I=\frac{9}{250}\int_{0}^{\frac{\pi}{2}}1-\cos{2\theta}\ d\theta}\\ \displaystyle{I=\frac{9}{250}\left[\theta-\frac{1}{2}\sin{2\theta}\right]_{0}^{\frac{\pi}{2}}}\\ \displaystyle{I=\frac{9}{250}\left(\left(\frac{\pi}{2}-\frac{1}{2}\sin{\left(2\times\frac{\pi}{2}\right)}\right)-(0)\right)}\\ \displaystyle{I=\frac{9\pi}{250\times2}-0}\\ \displaystyle{I=\frac{9\pi}{500}}\\$