antiderivative around inverse trigonometric function

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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?

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2 Answers

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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.

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

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