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    Not sure if I'm missing something simple here, but the differential of arcsin and arccos are the same apart from the minus sign with the arccoss one, right?

    So if you integrated the differential of arcsin, you get arcsin...
    What if you integrated the differential of arccos, took the -1 out, then got an answer of -arcsin?

    Is this wrong? Because -arcsin doesnt equal arccos, does it?

    What am I missing or doing wrong? Thank you to all maths nerds out there! Sorry if this is too obscure to get an answer, I shall continue my scouring of the web.
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    (Original post by physicsfuntimes)
    Not sure if I'm missing something simple here, but the differential of arcsin and arccos are the same apart from the minus sign with the arccoss one, right?

    So if you integrated the differential of arcsin, you get arcsin...
    What if you integrated the differential of arccos, took the -1 out, then got an answer of -arcsin?

    Is this wrong? Because -arcsin doesnt equal arccos, does it?

    What am I missing or doing wrong? Thank you to all maths nerds out there! Sorry if this is too obscure to get an answer, I shall continue my scouring of the web.
    The arbitrary constants of integration are very important as they are different, depending on whether you take the -1 out or not.

    If you don't see how this comes about, recall that \arcsin x + \arccos x = \frac{\pi}{2}, so that \arccos x = \frac{\pi}{2} - \arcsin x. The only difference between the LHS and the RHS is a constant, which would be reflected in the different constants of integration.
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    You're doing nothing wrong. In fact, \arcsin x + \arccos x \equiv \dfrac{\pi}{2}, and so differentiating both sides we must get \dfrac{d}{dx} \arcsin x + \dfrac{d}{dx} \arccos x = 0, as you observe.

    EDIT: Beaten to it! Damn you Farhan... :p:
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    (Original post by nuodai)
    You're doing nothing wrong. In fact, \arcsin x + \arccos x \equiv \dfrac{\pi}{2}, and so differentiating both sides we must get \dfrac{d}{dx} \arcsin x + \dfrac{d}{dx} \arccos x = 0, as you observe.

    EDIT: Beaten to it! Damn you Farhan... :p:
    I knew it wouldn't be long before you posted on the matter. :p:
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    (Original post by nuodai)
    You're doing nothing wrong. In fact, \arcsin x + \arccos x \equiv \dfrac{\pi}{2}, and so differentiating both sides we must get \dfrac{d}{dx} \arcsin x + \dfrac{d}{dx} \arccos x = 0, as you observe.

    EDIT: Beaten to it! Damn you Farhan... :p:

    (Original post by Farhan.Hanif93)
    I knew it wouldn't be long before you posted on the matter. :p:
    In fairness you did write a very detailed post on the subject earlier nuodai, only fair to share
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    Haha cheers guys, now I know who to go to
 
 
 
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