Integral : $\int \frac{\sqrt{9-x^2}}{x^2}$

Question

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 ?

Answer 1

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.

Answer 2

$$ \sin{x}=\cos{(\pi/2-x)}, $$ 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.