evaluate the integral.\n int_{0}^{2sqrt{2}} \frac{x^{2}}{sqrt{16 - x^{2}}} dx=square \text{(type an exact…

Explanation:

Step1: Use trigonometric substitution

Let $x = 4\sin\theta$, then $dx=4\cos\theta d\theta$. When $x = 0$, $\theta=0$; when $x = 2\sqrt{2}$, $\sin\theta=\frac{\sqrt{2}}{2}$, so $\theta=\frac{\pi}{4}$. Substitute into the integral: [ \begin{align*} \int_{0}^{2\sqrt{2}}\frac{x^{2}}{\sqrt{16 - x^{2}}}dx&=\int_{0}^{\frac{\pi}{4}}\frac{(4\sin\theta)^{2}}{\sqrt{16-16\sin^{2}\theta}}\cdot4\cos\theta d\theta\ &=\int_{0}^{\frac{\pi}{4}}\frac{16\sin^{2}\theta}{\sqrt{16(1 - \sin^{2}\theta)}}\cdot4\cos\theta d\theta\ &=\int_{0}^{\frac{\pi}{4}}\frac{16\sin^{2}\theta}{4\cos\theta}\cdot4\cos\theta d\theta\ &=16\int_{0}^{\frac{\pi}{4}}\sin^{2}\theta d\theta \end{align*} ]

Step2: Use the double - angle formula

Recall that $\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}$. Then $16\int_{0}^{\frac{\pi}{4}}\sin^{2}\theta d\theta=16\int_{0}^{\frac{\pi}{4}}\frac{1 - \cos(2\theta)}{2}d\theta$. [ \begin{align*} 16\int_{0}^{\frac{\pi}{4}}\frac{1 - \cos(2\theta)}{2}d\theta&=8\int_{0}^{\frac{\pi}{4}}(1-\cos(2\theta))d\theta\ &=8\left[\theta-\frac{1}{2}\sin(2\theta)\right]_{0}^{\frac{\pi}{4}} \end{align*} ]

Step3: Evaluate the definite integral

[ \begin{align*} 8\left[\theta-\frac{1}{2}\sin(2\theta)\right]_{0}^{\frac{\pi}{4}}&=8\left(\frac{\pi}{4}-\frac{1}{2}\sin\left(\frac{\pi}{2}\right)\right)-8(0 - \frac{1}{2}\sin(0))\ &=8\left(\frac{\pi}{4}-\frac{1}{2}\times1\right)-0\ &=2\pi - 4 \end{align*} ]

Answer:

$2\pi - 4$