Integral : $\int \frac{\sqrt{9-x^2}}{x^2}$ I have the following integral : 
$$\int \frac{\sqrt{9-x^2}}{x^2} = -\frac{\sqrt{9-x^2}}{x} - \arccos (\frac{x}{3})$$
But it seems like I must use $\arcsin$... Is there a difference between the two ?
 A: $$ \sin{x}=\cos{(\pi/2-x)}, $$
so with the usual conventions,
$$ \arccos{x} = \frac{\pi}{2}-\arcsin{x}. $$
Therefore from the point of view of integration, there is very little difference between them:
$$ \arcsin'{x} = \frac{1}{\sqrt{1-x^2}} = -\arccos'{x}, $$
so the substitutions are basically identical up to constants.
A: First off the sign on the $\arccos(\frac{x}{3})$ is wrong.
I guess you could use the constant of integration to solve this. From the definitions of the inverse trig functions we know that 
$\arccos x = \frac{\pi}{2} - \arcsin x$
So your antiderivative can be written as
\begin{equation}
-\frac{\sqrt{9 - x^2}}{x^2} + \arccos(\frac{x}{3}) =  -\frac{\sqrt{9 - x^2}}{x^2} + \frac{\pi}{2} - \arcsin(\frac{x}{3})
\end{equation}
Since $-\frac{\sqrt{9 - x^2}}{x^2} + \frac{\pi}{2} - \arcsin(\frac{x}{3})$ and $-\frac{\sqrt{9 - x^2}}{x^2} - \arcsin(\frac{x}{3})$ have the same derivative(the constant $\frac{\pi}{2}$ does nothing) it means that they are both antiderivatives of your original function. This is why a $ + C$ is added at the end of the antiderivative's general form. 
