Answer
$$\int^{1}_{1/\sqrt3}\frac{dy}{y\sqrt{4y^2-1}}=\frac{\pi}{6}$$
Work Step by Step
$$A=\int^{1}_{1/\sqrt3}\frac{dy}{y\sqrt{4y^2-1}}=\int^{1}_{1/\sqrt3}\frac{dy}{y\sqrt{(2y)^2-1^2}}$$
Set $u=2y$, which means $$du=2dy$$ $$dy=\frac{1}{2}du$$
Also, we would have $y=1/2u$
For $y=1$, we have $u=2$ and for $y=1/\sqrt3$, we have $u=2/\sqrt3$
Therefore, $$A=\int^{2}_{2/\sqrt3}\frac{\frac{1}{2}du}{\frac{1}{2}u\sqrt{u^2-1^2}}=\int^{2}_{2/\sqrt3}\frac{du}{u\sqrt{u^2-1^2}}$$
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, $$A=\frac{1}{1}\sec^{-1}|u|\Big]^{2}_{2/\sqrt3}$$ $$A=\sec^{-1}2-\sec^{-1}\frac{2}{\sqrt3}$$ $$A=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$$
