Is this u-substitution valid? It is not necessary to take any example of any integral, so I'll just drop it: $$u=\sin x; \ \ \ du=\cos x dx$$ which my question is rather here:
Is it possible to do: (squaring both sides) $$u^2=\sin^2(x)=1-\cos^2(x)$$ $$\cos(x)=\sqrt{1-u^2}$$ is it? I don't think it's true for the simple fact that $\sqrt{x^2}=|x|$ and not just $x$ but my question earlier today talked about how domain can be somewhat dealt with the constant $+C$ as shown in my question:
My question about different domains on indefinite integrals
So I'm very curious on whether it would be valid or not
 A: Of course squaring still yields a true identity, but as your comment regarding absolute value suggests (and as Brian Moehring made explicit in the comments) $\cos x = \sqrt{1 - u^2}$ cannot hold on any interval where $\cos x < 0$.
As an example, consider
$$\int_0^\pi \sin x \cos^2 x \,dx .$$
The conventional substitution $v = \cos x$ quickly gives the value $\frac{2}{3}$.
On the other hand, taking $u = \sin x$, $du = \cos x\, dx$, and replacing $\cos x \rightsquigarrow \sqrt{1 - u^2}$ gives the incorrect value
$$\int_0^0 u \sqrt{1 - u^2} \,du = 0 .$$
To avoid this error, we can first split our domain of integration into subintervals on which $\cos x \geq 0$ or $\cos x \leq 0$, say, $[0, \frac{\pi}{2}]$ and $[\frac{\pi}{2}, \pi]$, whereon we thus have $\cos x = \sqrt{1 - u^2}$ and $\cos x = -\sqrt{1 - u^2}$, respectively. Then, our original integral becomes
\begin{align}
\int_0^\pi \sin x \cos^2 x \,dx
&= \int_0^\frac\pi2 \sin x \cos^2 x \,dx + \int_\frac\pi2^\pi \sin x \cos^2 x \,dx \\
&= \int_0^1 u \sqrt{1 - u^2} \,du + \int_1^0 u \left(-\sqrt{1 - u^2}\right) \,du \\
&= 2 \int_0^1 u \sqrt{1 - u^2} \,du \\
&=\frac23 .
\end{align}
