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,-\frac{\sqrt2}2\right)`, only apply to the original variable of integration (y).

When you make the substitution
`y=\frac12\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=\frac12\sec t\implies t=\arcsec(-2)=\frac{2\pi}3`.

When
`y=-\frac{\sqrt2}2`, you have
`-\frac{\sqrt2}2=\frac12\sec t\implies t=\arcsec(-\sqrt2)=\frac{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.