Answer
$$ - \frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^{ - \sqrt 2 /2} {\frac{{dy}}{{y\sqrt {4{y^2} - 1} }}} \cr
& = \int_{ - 1}^{ - \sqrt 2 /2} {\frac{{dy}}{{y\sqrt {{{\left( {2y} \right)}^2} - 1} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = 2y,{\text{ so that }}du = 2dy \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}y = - 2/\sqrt 2,{\text{ then }}u = 2\left( { - \sqrt 2 /2} \right) = - \sqrt 2 \cr
& \,\,\,\,\,\,{\text{If }}y = - 1,{\text{ then }}u = 2\left( { - 1} \right) = - 2 \cr
& {\text{then}} \cr
& \int_{ - 1}^{ - \sqrt 2 /2} {\frac{{dy}}{{y\sqrt {{{\left( {2y} \right)}^2} - 1} }}} = \int_{ - 2}^{ - 4\sqrt 2 } {\frac{{du/2}}{{y\sqrt {{u^2} - 1} }}} = \int_{ - 2}^{ - 4\sqrt 2 } {\frac{{du}}{{2y\sqrt {{u^2} - 1} }}} \cr
& = \int_{ - 2}^{ - \sqrt 2 } {\frac{{du}}{{u\sqrt {{u^2} - 1} }}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}se{c^{ - 1}}\left| {\frac{u}{a}} \right| + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = \left( {{{\sec }^{ - 1}}\left| u \right|} \right)_{ - 2}^{ - \sqrt 2 } + C \cr
& = {\sec ^{ - 1}}\left| { - \sqrt 2 } \right| - {\sec ^{ - 1}}\left| { - 2} \right| \cr
& = \frac{\pi }{4} - \frac{\pi }{3} \cr
& = - \frac{\pi }{{12}} \cr} $$
