antiderivative around inverse trigonometric function

Question

I'm just learning about antiderivative that resulted some inverse trigonometric functions, since derivative of $ \arccos (x)$ is $-\frac {1} {\sqrt{1-x^2}}$. I tried put this value back into integral calculator, but I got $- \arcsin (x)$ instead of $ \arccos (x)$. even though $-\arcsin (x) \neq \arccos (x)$ so why this happen? Kow should I fix my understanding?

Answer 1

$$ \arccos(x) = \frac{\pi }{2}-\arcsin(x)\implies (\arccos(x))' =-(\arcsin(x))' $$

Answer 2

That's because $$ \arcsin(x)+\arccos(x) = \frac{\pi }{2} $$ and $\pi/2$ is a constant.

When integrating, always remember that there's a constant $C$ in the result.