Evaluate $\int\frac{dx}{\sqrt{a^2-x^2}}$
My Attempt
Set $x=a\sin\theta\implies dx=a\cos\theta d\theta$
Thanx @cansomeonehelpmeout for the hint.
$\theta=\sin^{-1}\frac{x}{a}\implies\theta\in[-\pi/2,\pi/2]\implies|\cos\theta|=\cos\theta$ $$ \begin{align} \int\frac{dx}{\sqrt{a^2-x^2}}&=\int\frac{a\cos\theta d\theta}{\sqrt{a^2-a^2\sin^2\theta}}=\int\frac{a\cos\theta d\theta}{|a|\sqrt{1-\sin^2\theta}}=\int\frac{a.\cos\theta d\theta}{|a|\sqrt{\cos^2\theta}}\\ &=\int\frac{a.\cos\theta d\theta}{|a|.|\cos\theta|}=\int\frac{a.\cos\theta d\theta}{|a|.\cos\theta}=\begin{cases}\;\:\:\int d\theta=\theta+C \text{ if }a>0\\-\int d\theta= -\theta+C \text{ if }a<0\end{cases}\\ &=\begin{cases}\;\:\:\sin^{-1}\frac{x}{a}+C \text{ if }a>0\\-\sin^{-1}\frac{x}{a}+C \text{ if }a<0\end{cases} \end{align} $$
As the $\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+C$ is whats given in all references why am I getting this result ?
I can see that the function is independent of the sign of $a$ but why do I seem to get solutions dependent on the sign of $a$ ?
But my doubt is how do I eliminate the negative result while integrating the function ?
Note: I undestand $\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}\implies \frac{d}{dx}\sin^{-1}\frac{x}{a}=\frac{1}{a.\sqrt{1-\frac{x^2}{a^2}}}=\frac{1}{\sqrt{a^2-x^2}}$
It's hard to see when there are no bounds to the integral but it might be because of that. Sometimes the change in the bounds in a substitution bring a change in the sign, that might alter the result.
Anyway, here there is a way to integrate it without a substitution (I prefer to avoid them when there are no bounds in the integral, since you can't always verify that it's bijective):
$$\int \frac{dx}{\sqrt{a^2 -x^2}} = \frac{1}{|a|}\int \frac{dx}{\sqrt{1-\frac{x^2}{a^2}}} = \sin^{-1}(x/a) + C$$
Edit: oh no my bad actually you're right it's what you found. Since there is a |a| that appears. The books might just be considering the case a>0 I think