Answer
$$\int^{2/3}_{\sqrt2/3}\frac{dy}{|y|\sqrt{9y^2-1}}=\frac{\pi}{12}$$
Work Step by Step
$$A=\int^{2/3}_{\sqrt2/3}\frac{dy}{|y|\sqrt{9y^2-1}}=\int^{2/3}_{\sqrt2/3}\frac{dy}{|y|\sqrt{(3y)^2-1^2}}$$
Since here we consider $y\in[\sqrt2/3,2/3]$, we know that $y\gt0$, so $|y|=y$
Set $u=3y$, which means $$du=3dy$$ $$dy=\frac{1}{3}du$$
Also, we would have $y=1/3u$
For $y=2/3$, we have $u=2$ and for $y=\sqrt2/3$, we have $u=\sqrt2$
Therefore, $$A=\int^{2}_{\sqrt2}\frac{\frac{1}{3}du}{\frac{1}{3}u\sqrt{u^2-1^2}}=\int^{2}_{\sqrt2}\frac{du}{u\sqrt{u^2-1^2}}$$
We have $$\int\frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}\sec^{-1}\Big|\Big(\frac{x}{a}\Big)\Big|+C$$
Therefore, $$A=\frac{1}{1}\sec^{-1}|u|\Big]^2_{\sqrt2}$$ $$A=\sec^{-1}2-\sec^{-1}\sqrt2$$ $$A=\frac{\pi}{3}-\frac{\pi}{4}$$ $$A=\frac{\pi}{12}$$
