I am trying to find the integral: $$\int \frac{\cos^2(x)}{\sqrt{\cos^4(x) + \cos^2(x) + 1}} \, dx$$
Which I found here: http://www.12000.org/my_notes/ten_hard_integrals/index.htm
I start by substituting $x = \arccos(u)$, making $dx = \frac{-1}{\sqrt{1-u^2}} \, du$. This means the integral is equal to:
$$-\int \frac{u^2}{\sqrt{(u^4+u^2+1)(1 - u^2)}} \, du$$
for any $x$. The bottom half is equal to $1-u^6$, so if $v = u^3$, then the integral becomes:
$$-\frac 1 3 \int \frac{1}{\sqrt{1-v^2}} \, dv$$
Or:
$$-\frac 1 3 \arcsin(\cos^3(x)) + C$$
The problem I have, though, is that the derivative of this function is not equal to the function I integrated. Wolfram Alpha shows that it is only true for $x \in [0 + 2\pi n, \pi + 2\pi n]$ for any integer $n$, and that the real integral is the absolute value of the one I found.
But where did this restriction come from? I don't see any step where I did something that isn't true for all $x$'s.
The range of $\arccos$ is $[0,\pi].$ So when you make the substitution $x = \arccos u,$ you are literally saying that $0 \leq x \leq \pi.$
Fortunately, thanks to the fact that $\cos(x + 2\pi) = \cos(x),$ you end up with indefinite integral that also works for $[2\pi,3\pi],$ $[4\pi,5\pi],$ and so forth. But to fill in the "missing" values of $x$ you need something like $x = -\arccos u,$ which has the range $[-\pi,0].$