Just to go into more detail than I did in our PMs and the comments on your last question...
You have to keep in mind that the limits of integration, the interval \(\left(-1,-\dfrac{\sqrt2}2\right)\), only apply to the original variable of integration (y).
When you make the substitution \(y=\dfrac12\sec t\), you not only change the variable but also its domain. To find out what the new domain is is a matter of plugging in every value in the y-interval into the substitution relation to find the new t-interval domain for the new variable (t).
After replacing \(y\) and the differential \(\mathrm dy\) with the new variable \(t\) and differential \(\mathrm dt\), you saw that you could reduce the integral to -1. This is a continuous function, so the new domain can be constructed just by considering the endpoints of the y-interval and transforming them into the t-domain.
When \(y=-1\), you have \(-1=\dfrac12\sec t\implies t=\arcsec(-2)=\dfrac{2\pi}3\).
When \(y=-\dfrac{\sqrt2}2\), you have \(-\dfrac{\sqrt2}2=\dfrac12\sec t\implies t=\arcsec(-\sqrt2)=\dfrac{3\pi}4\).
Geometrically, this substitution allows you to transform the area as in the image below. Naturally it's a lot easier to find the area under the curve in the second graph than it is in the first.
