TITLE: Why does the tan inverse integral have a 1/a but not the sin inverse one?
QUESTION [6 upvotes]: $$\begin{align*}
\int \frac{du}{\sqrt{a^2-u^2}} &= \sin^{-1}\left(\frac{u}{a}\right) + K\\
\int \frac{du}{a^2+u^2} &= \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + K \end{align*}$$
Always wondered why but I had no idea.

REPLY [9 votes]: 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.