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Math Help - Why the indefinite integral of the derivative of arccos(x) is not equal to arccos(x)

  1. #1
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    Smile 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.
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  2. #2
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    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.
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  3. #3
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    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.
    Thanks from Melsi
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  4. #4
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    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!
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