evaluate $int_{0}^{\frac{2}{5}}\frac{x^{2}}{sqrt{4 - 25x^{2}}}dx$.

Explanation:

Step1: Use substitution

Let $x = \frac{2}{5}\sin\theta$, then $dx=\frac{2}{5}\cos\theta d\theta$. When $x = 0$, $\theta=0$; when $x=\frac{2}{5}$, $\theta=\frac{\pi}{2}$. And $\sqrt{4 - 25x^{2}}=\sqrt{4-4\sin^{2}\theta}=2\cos\theta$. The integral becomes $\int_{0}^{\frac{\pi}{2}}\frac{(\frac{2}{5}\sin\theta)^{2}}{2\cos\theta}\cdot\frac{2}{5}\cos\theta d\theta=\frac{8}{125}\int_{0}^{\frac{\pi}{2}}\sin^{2}\theta d\theta$.

Step2: Use the identity $\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}$

We have $\frac{8}{125}\int_{0}^{\frac{\pi}{2}}\frac{1 - \cos(2\theta)}{2}d\theta=\frac{4}{125}\int_{0}^{\frac{\pi}{2}}(1-\cos(2\theta))d\theta$.

Step3: Integrate term - by - term

$\frac{4}{125}\left[\int_{0}^{\frac{\pi}{2}}1d\theta-\int_{0}^{\frac{\pi}{2}}\cos(2\theta)d\theta\right]$. The integral of $1$ with respect to $\theta$ is $\theta$, and $\int\cos(2\theta)d\theta=\frac{1}{2}\sin(2\theta)$. So $\frac{4}{125}\left[\theta-\frac{1}{2}\sin(2\theta)\right]_{0}^{\frac{\pi}{2}}$.

Step4: Evaluate the definite integral

$\frac{4}{125}\left[\left(\frac{\pi}{2}-\frac{1}{2}\sin(\pi)\right)-\left(0 - \frac{1}{2}\sin(0)\right)\right]=\frac{2\pi}{125}$.

Answer:

$\frac{2\pi}{125}$