Question

Why did he let (y) equals to (1/2 sec t) ? And how did (-1) & (- √2/2) become (2π/3) & (3π/4) ?!!

Answer 1

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.