When we substitute $x=\sin(t)$ into $$\int \sqrt{1-x^2}dx,$$ why we only take the "$+$" sign even though $\sqrt{1-\sin^2(t)}=\pm \cos(t)\;?$
EDIT: Thank you to @Cameron Williams for the suggested problem. I know for the definite integral, we need to discuss carefully according to the quadrant for + or -, and that's no ambiguity. But here I want to ask for the indefinite case, where there is no specified quadrant, then why we only take the "+" sign?
EDIT: Thank you to @EeveeTrainer for the suggested problem. But here it is not exactly to take square root on a number, but a function. $\sqrt{4}=2$ is not the same as $\sqrt{\cos^2(x)}=\cos(x)$. Because $\cos(x)$ might be negative, depends on the quadrant of $x$ previously specified. Is the underneath assumption to choose a quadrant such that both $\sin$ and $\cos$ are positive, even though there is no quadrant specified for an indefinite integral?
EDIT: Thank you to @finch's comment on the conventional interval chosen for $\sin(x)$ (or $\cos(x)$) so that they are invertible. For $\sin(x), ~~[-\frac{\pi}2, \frac{\pi}2];$ for $\cos(x), ~~[0, \pi].$ So, if I want to use substitute $$x=\sin(\theta)$$ into $$I_2=\int \sqrt{x^2}\sqrt{1-x^2} dx,$$ which interval should I choose? Do I choose $$[-\frac{\pi}2, \frac{\pi}2]\cap [0, \pi] ~?$$
ADDENDUM
Say, I choose the domain for $\theta $ as $$[-\frac{\pi}2, \frac{\pi}2].$$ Then the integral becomes $$I_2=\int \sqrt{\sin^2(\theta)}\sqrt{\cos^2(\theta)} \cos(\theta)~d\theta=\int |\sin(\theta)|\cos(\theta) \cos(\theta)d\theta$$
$$\begin{align}
\text{If}~~~ \theta\in[-\frac{\pi}{2},0],~~I_2&=\frac{1}{3}\cos^3(\theta)+C=\frac{1}{3}(1-x^2)^{3/2}+C\\
\\
\text{If} ~~~\theta\in(0,\frac{\pi}{2}],~~I_2&=-\frac{1}{3}\cos^3(\theta)+C=-\frac{1}{3}(1-x^2)^{3/2}+C
\end{align}$$
Why do we usually choose only the bottom solution?
Best Answer
Assuming $t\in[\pi/2, 3\pi/2]$ is not a problem. Let $x=\sin t$, but $t = \color{red}{\pi-}\sin^{-1}x$, $$\begin{align*} \int\sqrt{1-x^2}dx &= \int\sqrt{1-\sin^2t}\cdot \cos t\ dt\\ &=\int-\cos^2t\ dt\\ &=\int-\frac{1+\cos2t}{2}dt\\ &=-\frac t2-\frac{\sin 2t}4+C\\ &=-\frac t2-\frac{\sin t\cos t}2+C\\ &= -\frac{\pi-\sin^{-1}x}2 -\frac{x\left(\color{red}-\sqrt{1-x^2}\right)}2+C\\ &= \frac{\sin^{-1}x + x\sqrt{1-x^2}}{2} + C' \end{align*}$$
Just be careful when representing $t$ and $\cos t$ in terms of $x$.
For the ADDENDUM "why do we usually only choose the bottom solution?" No, for example from WolframAlpha, the indefinite integral assuming $x$ is real also depends on the sign of $x$:
$$I_2 = -\frac{\left(1-x^2\right)^{3/2}}{3\operatorname {sgn}(x)} + C$$
which means when $x=\sin\theta$ and $\theta$ are both negative, this matches your first form of $I_2$.