Role of the absolute value in $\int \frac{dx}{\sqrt{1-x^2}}$ In the derivation of the value of the indefinite integral \begin{equation} \int \frac{dx}{\sqrt{1-x^2}}, \end{equation} I can substitute $x = \sin(u)$, $dx = \cos(u)du$ to get this: \begin{equation} \int \frac{\cos(u)du}{\sqrt{1-\sin^2(u)}} = \int \frac{\cos(u)du}{\lvert \cos(u)\rvert} \overset{?}{=} \int du = u = \arcsin(x), \end{equation} But I'm sceptical about the division of the regular cosine over the absolute value of the cosine...

$$\dfrac{d(\sin u)}{du}=\cos u$$

If $u=\arcsin x,\sin u=x$ and $\dfrac\pi2\le u\le\dfrac\pi2\implies\cos u\ge0$

and consequently, $\cos u=+\sqrt{1-x^2}$

and $$\dfrac{d(\arcsin x)}{dx}=\dfrac{du}{d(\sin u)}=\dfrac1{\cos u}=?$$

But we know $$\displaystyle\dfrac{dy}{\sqrt{a^2-y^2}}=\dfrac1{|a|}\arcsin\dfrac xa+K$$