Evaluate the integral. \[ \int_{0}^{0.8} \frac{x^{2}}{\sqrt{16-25 x^{2}}} d x \]

To evaluate the integral \[ I = \int_{0}^{0.8} \frac{x^{2}}{\sqrt{16-25 x^{2}}} \, dx, \] we can use the substitution \(x = \frac{4}{5} \sin(\theta)\). This gives us the following transformations: - When \(x = 0\), \(\theta = 0\). - When \(x = 0.8\), \(x = \frac{4}{5} \sin(\theta)\) gives \(\sin(\theta) = 1\), so \(\theta = \frac{\pi}{2}\). Next, we differentiate to find \(dx\): \[ dx = \frac{4}{5} \cos(\theta) \, d\theta. \] Now, we calculate the integral in terms of \(\theta\): - Since \(x^2 = \left(\frac{4}{5} \sin(\theta)\right)^2 = \frac{16}{25} \sin^2(\theta)\), - And for the term under the square root, we have: \[ 16 - 25x^2 = 16 - 25\left(\frac{4}{5} \sin(\theta)\right)^2 = 16 - 25 \cdot \frac{16}{25} \sin^2(\theta) = 16(1 - \sin^2(\theta)) = 16 \cos^2(\theta). \] Thus, \(\sqrt{16 - 25x^2} = 4 \cos(\theta)\). Now substitute these into the integral: \[ I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{16}{25} \sin^2(\theta)}{4 \cos(\theta)} \cdot \frac{4}{5} \cos(\theta) \, d\theta. \] The \(\cos(\theta)\) terms cancel out: \[ I = \int_{0}^{\frac{\pi}{2}} \frac{16}{25} \sin^2(\theta) \cdot \frac{4}{5} \, d\theta = \frac{64}{125} \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta. \] We can calculate \(\int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta\) using the identity \[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}, \] which gives: \[ \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos(2\theta)) \, d\theta = \frac{1}{2} \left[ \theta - \frac{1}{2} \sin(2\theta) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}. \] Now substituting this back into the integral \(I\): \[ I = \frac{64}{125} \cdot \frac{\pi}{4} = \frac{16\pi}{125}. \] Thus, the value of the integral is \[ \boxed{\frac{16\pi}{125}}. \]