Why does the tan inverse integral have a 1/a but not the sin inverse one?

I think the most intuitive way to think of this is in terms of dimensional analysis.

Since $u^2$ and $a^2$ are added, they must have the same dimensions, so $u$ and $a$ must as well. Say, for specificity, that $u$ and $a$ are both lengths.

Then the left integrand is dimensionless (as $du$ and $\sqrt{a^2-u^2}$ are both lengths), so it can integrate to a pure number. But the right integrand has units of inverse length (as $a^2+u^2$ is an area). So it must integrate to some quantity which also has units of inverse length; as $a$ is the only constant around with units of length, it's natural to write this as $\frac{1}{a}$ times some function of $u$ which is a pure number.

For the latter, you have $\dfrac{1}{a}\cdot d\left(\dfrac{u}{a}\right)= \dfrac{du}{a^2}$ while the former you have $\dfrac{du}{a} = d\left(\dfrac{u}{a}\right)$. The $\dfrac{1}{a}$ shows up to balance the $a^2$ at the denominator.