\(\displaystyle\int_{-1}^{-\sqrt2/2}\frac{\mathrm dy}{y\sqrt{4y^2-1}}\)
Try setting \(y=\dfrac12\sec t\), so that \(\mathrm dy=\dfrac12\sec t\tan t\,\mathrm dt\). The interval then changes from \(-1\le y\le-\dfrac{\sqrt2}2\) to \(\dfrac{2\pi}3\le t\le \dfrac{3\pi}4\).
Now, the integral is
\(\displaystyle \int_{2\pi/3}^{3\pi/4} \frac{\dfrac12\sec t\tan t}{\dfrac12\sec t\sqrt{4\left(\dfrac12\sec t\right)^2-1}}\,\mathrm dt=\int_{2\pi/3}^{3\pi/4}\frac{\tan t}{\sqrt{\sec^2t-1}}\,\mathrm dt\)
The Pythagorean identity lets you reduce the denominator:
\(\sqrt{\sec^2t-1}=\sqrt{\tan^2t}=|\tan t|\)
Since \(\tan t<0\) for the given interval, you have \(|\tan t|=-\tan t\), which means the integral is equal to
\(\displaystyle\int_{2\pi/3}^{3\pi/4}\frac{\tan t}{-\tan t}\,\mathrm dt=-\int_{2\pi/3}^{3\pi/4}\mathrm dt=-\dfrac\pi{12}\)
