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Trigonometric functions and integrals

  1. Dec 7, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm searching for the integral that gives arcosu

    2. Relevant equations
    as we know : ∫u'/[1-u^2]^0.5 dx = arcsinu
    derivative of arccosu = -u'/[1-u^2]^0.5 + C
    derivative of arcsinu= u'/[1-u^2]^0.5
    3. The attempt at a solution
    when I type the -u'/[1-u^2]^0.5 on the online integral calculator it always gives me -arcsinu +C
    Why it doesn't gives arccosu
    is the arccosu=-arcsinu?!
    and what is the integral that gives arcosu?
     
  2. jcsd
  3. Dec 7, 2016 #2

    cnh1995

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    cos-1u= -sin-1u+90°.
    Sine and cosine functions have a phase difference of 90°.
     
  4. Dec 7, 2016 #3

    Ray Vickson

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    No, they differ by ##\pi/2## radians. It is a great mistake to mix trigonometric integrals with degree measurements; the calculus of trig functions is in radian units only!
     
  5. Dec 7, 2016 #4

    Mark44

    Staff: Mentor

    @Any Help, please post questions about calculus in the Calculus & Beyond section, not the Precalc section. I have moved your thread.
     
  6. Dec 7, 2016 #5
    then what's the integral that gives arccos u?
     
  7. Dec 7, 2016 #6

    Ray Vickson

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    There are two antiderivatives of ##1/\sqrt{1-x^2}##; they are ##\arcsin(x)+C## and ##-\arccos(x)+C##. Typically, sources like tables of integrals, or computer algebra packages, give only ONE of the two, and those seem to always be the ##\arcsin(x)## version. When you do it manually you are certainly allowed to give ##-\arccos(x) + C## as an answer; it is absolutely correct, because it is the same as ##\arcsin(x) + K## for some constant ##K## that is different from the constant ##C##.
     
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