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 [tex]\left(-1,-\dfrac{\sqrt2}2\right)[/tex], only apply to the original variable of integration (y).
When you make the substitution [tex]y=\dfrac12\sec t[/tex], 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 [tex]y[/tex] and the differential [tex]\mathrm dy[/tex] with the new variable [tex]t[/tex] and differential [tex]\mathrm dt[/tex], 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 [tex]y=-1[/tex], you have [tex]-1=\dfrac12\sec t\implies t=\arcsec(-2)=\dfrac{2\pi}3[/tex].
When [tex]y=-\dfrac{\sqrt2}2[/tex], you have [tex]-\dfrac{\sqrt2}2=\dfrac12\sec t\implies t=\arcsec(-\sqrt2)=\dfrac{3\pi}4[/tex].
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.
