evaluate ∫₀^(4/3) x²/√(16 - 9x²) dx.

Explanation:

Step1: Use substitution

Let $x = \frac{4}{3}\sin\theta$, then $dx=\frac{4}{3}\cos\theta d\theta$. When $x = 0$, $\theta=0$; when $x=\frac{4}{3}$, $\theta=\frac{\pi}{2}$. And $\sqrt{16 - 9x^{2}}=\sqrt{16-16\sin^{2}\theta}=4\cos\theta$, $x^{2}=\frac{16}{9}\sin^{2}\theta$.

Step2: Rewrite the integral

The integral $\int_{0}^{\frac{4}{3}}\frac{x^{2}}{\sqrt{16 - 9x^{2}}}dx$ becomes $\int_{0}^{\frac{\pi}{2}}\frac{\frac{16}{9}\sin^{2}\theta}{4\cos\theta}\cdot\frac{4}{3}\cos\theta d\theta=\frac{64}{81}\int_{0}^{\frac{\pi}{2}}\sin^{2}\theta d\theta$.

Step3: Use the identity $\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}$

$\frac{64}{81}\int_{0}^{\frac{\pi}{2}}\sin^{2}\theta d\theta=\frac{64}{81}\int_{0}^{\frac{\pi}{2}}\frac{1-\cos(2\theta)}{2}d\theta=\frac{32}{81}\int_{0}^{\frac{\pi}{2}}(1 - \cos(2\theta))d\theta$.

Step4: Integrate term - by - term

$\frac{32}{81}\left[\int_{0}^{\frac{\pi}{2}}1d\theta-\int_{0}^{\frac{\pi}{2}}\cos(2\theta)d\theta\right]$. The integral of $1$ with respect to $\theta$ is $\theta$, and for $\int\cos(2\theta)d\theta=\frac{1}{2}\sin(2\theta)$. So we have $\frac{32}{81}\left[\theta-\frac{1}{2}\sin(2\theta)\right]_{0}^{\frac{\pi}{2}}$.

Step5: Evaluate the definite integral

$\frac{32}{81}\left[\left(\frac{\pi}{2}-\frac{1}{2}\sin(\pi)\right)-\left(0 - \frac{1}{2}\sin(0)\right)\right]=\frac{16\pi}{81}$.

Answer:

$\frac{16\pi}{81}$