# Thread: Why the indefinite integral of the derivative of arccos(x) is not equal to arccos(x)

1. ## Why the indefinite integral of the derivative of arccos(x) is not equal to arccos(x)

Hello everyone,

I am checking with an online calculator some derivatives and integrals but there is a specific result I don't understand.

In particular, for the derivative of arccos(x) I get this result:

$\displaystyle \frac{\mathrm{d} (arccos(x))}{\mathrm{d} x} = -1/ \sqrt{1-x^2}$

which is fine.

However when I ask what is the integral of derivative of arccos I get in particualar:

$\displaystyle \int -1/\sqrt{1-x^2}dx = -sin^{-1}(x)$

I was sure that the calculator would give arccos(x) as an answer, but it didn't. I suspect that there is no function whose indefinite integral equals arccos(x), and I should bound the integral but I am not so sure about it.

The answer to this could be long and I don't want to bother anyone, so any reference keywords for me to search in google would be also fine.

Also this question might be incomplete or not clear, so I am here to provide any other information if needed.

2. ## Re: Why the indefinite integral of the derivative of arccos(x) is not equal to arccos

Answear:

Since the derivative of arccos(x) and arcsin(x) are equal by absolute value (the deffer in the negative sign only), then both arccos and arcsine can be used as a result of integration as soon as the proper sign is placed in front.

3. ## Re: Why the indefinite integral of the derivative of arccos(x) is not equal to arccos

Or you can notice that since:

$\displaystyle \arccos(x)+\arcsin(x)=\frac{\pi}{2},$

the answer it has given you for the indefinite integral is equal to $\displaystyle \arccos(x)$ up to an additive constant.

4. ## Re: Why the indefinite integral of the derivative of arccos(x) is not equal to arccos

That's a good point! Never crossed my mind!