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

• Mar 31st 2012, 07:30 AM
Melsi
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:

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

which is fine.

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

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

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.
• Mar 31st 2012, 08:28 AM
Melsi
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.
• Mar 31st 2012, 01:42 PM
Waldo750
Re: Why the indefinite integral of the derivative of arccos(x) is not equal to arccos
Or you can notice that since:

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

the answer it has given you for the indefinite integral is equal to $\arccos(x)$ up to an additive constant.
• May 3rd 2012, 11:45 PM
Melsi
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!