Answer :

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[tex]\displaystyle\int_{-1}^{-\sqrt2/2}\frac{\mathrm dy}{y\sqrt{4y^2-1}}[/tex]

Try setting [tex]y=\dfrac12\sec t[/tex], so that [tex]\mathrm dy=\dfrac12\sec t\tan t\,\mathrm dt[/tex]. The interval then changes from [tex]-1\le y\le-\dfrac{\sqrt2}2[/tex] to [tex]\dfrac{2\pi}3\le t\le \dfrac{3\pi}4[/tex].

Now, the integral is

[tex]\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[/tex]

The Pythagorean identity lets you reduce the denominator:

[tex]\sqrt{\sec^2t-1}=\sqrt{\tan^2t}=|\tan t|[/tex]

Since [tex]\tan t<0[/tex] for the given interval, you have [tex]|\tan t|=-\tan t[/tex], which means the integral is equal to

[tex]\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}[/tex]

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