# $\int\frac{1}{\sqrt{a^2-x^2}}dx\overset{?}=\cos^{-1}\left(\frac{x}{a}\right)$ [duplicate]

I'm trying to understand what's happened here. From what I've found online it should be $\sin^{-1}$ not $\cos^{-1}$. But then the proof doesn't work. I'm probably just missing something...

I'm assuming that the equation in the picture isn't of the form suggested in the title or something?

• Start with let $x=a\cos \theta$. Now, try. – user249332 Mar 20 '16 at 22:24
• Look at the picture I've linked, it's a bit more complicated haha, I'm just guessing that it's of the form in the title... – CanofDrink Mar 20 '16 at 22:26
• There $x=\rho-1/r_0$ and $a=e/r_0$. – user249332 Mar 20 '16 at 22:29
• Yeah I can see that, but from what I've seen online it says that it should be arcsin...but radio.astro.gla.ac.uk/a1dynamics/ellproof.pdf says it's arccos. – CanofDrink Mar 20 '16 at 22:31
• Here I am using the identity $$\arccos x+\arcsin x=\frac \pi 2.$$ – user249332 Mar 20 '16 at 22:37

When you do integration and find antiderivative you can only find a unique solution up to addition of constant. For example $\int 2xdx=x^2+C$, it's also true to write $\int 2xdx=x^2+1+C$, $\int 2xdx=x^2-100+C$, $\int 2xdx=x^2+\pi+C$, etc. because $C$ is an arbitrary constant that can "absorb" all the constant terms.
$\arcsin(x)+\arccos(x)=\frac{\pi}{2}\implies \arccos(x)=-\arcsin(x) +\frac{\pi}{2}$. Therefore if it's correct that $\int f(x) dx=-\arcsin(x)+C$ it's also correct that $\int f(x) dx=-\arcsin(x)+\frac{\pi}{2}+C=\arccos(x)+C$.
Assuming you haven't read any book's solution manual about this problem, we can start again by letting $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2}), x = a\sin \theta\Rightarrow dx = a\cos \theta d\theta\Rightarrow \sqrt{a^2-x^2} = \sqrt{a^2-a^2\sin^2\theta}= a\cos \theta\Rightarrow I = \displaystyle \int 1d\theta= \theta + C = \sin^{-1}\left(\dfrac{x}{a}\right)+C$