Compute $\int\frac{du}{\sqrt{a^2-u^2}}$ Please, can you explain why isn't the $$\int\frac{du}{\sqrt{a^2-u^2}} = \int\frac{du}{a\sqrt{1-(u/a)^2}}=\frac1a\sin^{-1}\bigg(\frac{u}{a}\bigg)+c$$ if $a > 0$ is a positive constant.

Let $v = \dfrac ua \implies u = a\cdot v$, and $\mathrm du = a\cdot\mathrm dv$.
Substitute in original expression,
$$\require{cancel}\begin{align}\int\dfrac{\mathrm du}{\sqrt{a^2 - u^2}} &\equiv \int\dfrac{a\cdot\mathrm dv}{\sqrt{a^2 - a^2v^2}}\\ &= \int\dfrac{a\cdot\mathrm dv}{a\sqrt{1 - v^2}}\\ &= \int\dfrac{\cancel{a}\cdot\mathrm dv}{\cancel{a}\sqrt{1 - v^2}}\\ &= \int\dfrac{\mathrm dv}{\sqrt{1 - v^2}} = \sin^{-1}\left(v\right) + C\end{align}$$
Reverse substitution,
$$\sin^{-1}\left(v\right) + C \equiv \sin^{-1}\left(\dfrac ua\right) + C$$
What you are missing in your calculation is that
$$\dfrac {\mathrm d}{\mathrm du}\sin^{-1}\left(\dfrac ua\right) = \dfrac 1a\cdot\dfrac1{\sqrt{1 - \left(u/a\right)^2}}$$
and, therefore, the $1/a$ is consumed as part of the integral.
