Why is arcsine the odd one out? Today in calc we learned these three identities:
$$\int{\frac{du}{\sqrt{a^2-u^2}}} =  \arcsin{\frac{u}{a}}+C$$
$$\int{\frac{du}{a^2+u^2}} =  \frac{1}{a}\arctan{\frac{u}{a}}+C$$
$$\int{\frac{du}{u\sqrt{u^2-a^2}}} =  \frac{1}{a}\arctan{\frac{|u|}{a}}+C$$
These identities are too complex for me to derive myself. I found it curious that the arcsine one was the only one without $\frac{1}{a}$. What's the reason for this?
 A: I think it might be useful to think of it in terms of the units the various quantities are in. If you've taken any Physics (or Chemistry), you know that whenever any two expressions are set equal to one another they must have the same units. For example $F = ma$ because the unit of force is mass times acceleration.
We can gain some insight as to why arcsine lacks the $1/a$ the same way. Let's say that $u$ and $a$ represent lengths (which is a good choice since we can imagine them denoting the sides of a right triangle), specifically meters. Whenever we take the arc-anything of something we get an angle back in units of radians (which, remember, is dimensionless).
If we look at the second equation you wrote, we can see that the right-hand side must have units of $1/\mathrm{meter}$ because it is an angle (which is dimensionless) divided by $a$ which is measured in meters. Does this match the left-hand side? Yes! Because the integrand is
$$ \frac{du}{a^2 + u^2} $$
which is $\mathrm{meter}/\mathrm{meter}^2 = 1/\mathrm{meter}$. The integral sign $\int$ simply means "add a bunch of these up" which doesn't alter the units.
Similarly, the third equation's right-hand side is also in units of $1/\mathrm{meter}$ for the same reason as before, and the left-hand side has units of$\newcommand{\meter}{\mathrm{meter}}$
$$ \frac{\meter}{\meter \cdot \sqrt{\meter^2}} = \frac{1}{\meter} $$
What about the first equation? There the left-hand side would have units of
$$ \frac{\meter}{\sqrt{\meter^2}} = 1 \leftarrow \textrm{dimensionless} $$
which matches the right-hand side whose units are radians (also dimensionless). If arcsine also had a $1/a$ factor, then the right-hand side would have units of $1/\meter$ which obviously cannot equal something that is dimensionless.
So the short answer for why the arcsine lacks the $1/a$ is that, in some sense, its integrand has different units than the other two.
A: Hint:
$$\frac1{\sqrt{a^2-u^2}}\sim\frac1{\sqrt{u^2}}=\frac1{u}\\\frac1{a^2+u^2}\sim\frac1{u^2}\\\frac1{u\sqrt{u^2-a^2}}\sim\frac1{u\sqrt{u^2}}=\frac1{u^2}$$
Thus, it is clear that the arcsin integral is off by a factor of $a$ compared to the others.
