We are given the following expressions that we eventually solve within an integral. These are the forms that we use trigonometric substitutions on:
$\sqrt{a^2 -x^2}$, $\sqrt{a^2 +x^2}$, and $\sqrt{x^2 -a^2}$.
The substitutions we are given are $x = a\sin(t)$, $x = a\tan(t)$, and $x = a\sec(t)$, respectively.
The problem tells us that after the substitutions, each of these expressions becomes $a \, \lvert \cos(t) \rvert$, $a \, \lvert \sec(t) \rvert$, and $a \, \lvert \tan(t) \rvert$, also respectively.
I'm tasked with finding the $t$-values that would render the absolute value signs useless. The question also gives the $t$-values for $\sqrt{x^2 - a^2}$.
$\sqrt{x^2 - a^2}$, $x = a \sec(t)$, $\sqrt{x^2 - a^2} = a \, \lvert \tan(t) \rvert$ for $0 < t \leq \frac{\pi}{2}$ or $\pi \leq t < \frac{3\pi}{2}$.
The $t$-values for the first range is listed as $\underline{\hspace{2em}} \leq t \leq \underline{\hspace{2em}}$, and the second is listed as $\underline{\hspace{2em}} < t < \underline{\hspace{2em}}$.
For some reason I cannot wrap my head around finding these $t$-values. I was thinking to look for the $t$-values where both the substitution function and inner-absolute value function are both positive, but that would mean the $t$-values for the second range should be the same $t$-values as the third range.
It's worth noting that when $a$ is a positive fixed number, we choose $t = \arcsin{\left(\frac{x}{a}\right)}$, or $t = \arctan{\left(\frac{x}{a}\right)}$, or $t = \operatorname{arcsec}{\left(\frac{x}{a}\right)}$, when it is convenient. This is so we don't have to worry about substitutions not being injective for integrals.
Let's use what we have above. As a head start, if we have the expression $\sqrt{x^2-a^2}$, then letting $t = \operatorname{arcsec}\left(\frac{x}{a}\right)$, or equivalently $x = a\sec{(t)}$, results in
$$\sqrt{x^2-a^2} = \sqrt{a^2\tan^2{(t)}} = \left|a\tan{(t)}\right|.$$
Now we need to get rid of the absolute values like what your directions say. Given that the tangent function is periodic, there would've been an infinite number of answers, but since $t$ is in the range of $\operatorname{arcsec}$, we get that
$$\sqrt{x^2-a^2} = \begin{cases} a\tan{(t)}, & x \geq a \text{ and } 0 \leq t < \frac{\pi}{2}\\ -a\tan{(t)}, & x \leq -a \text{ and } \frac{\pi}{2} < t \leq \pi. \end{cases}$$
I think you can do the rest of the assigned problems.