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Integral yielding 2 answers

  1. Dec 12, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\int \frac{1}{\sqrt{1-x^{2}}} dx[/itex]

    I seem to be getting two incompatible answers when I substitute x with [itex]sin \theta[/itex] and [itex]cos \theta[/itex]. Could someone please help me with where I'm going wrong.

    2. Relevant equations

    3. The attempt at a solution

    first attempt (the correct answer I believe):

    [itex]let\ x = sin \theta[/itex]

    [itex]dx = cos \theta\ d\theta[/itex]

    [itex]\int \frac{cos \theta}{\sqrt{1-sin^{2}\theta}} d\theta[/itex]

    [itex]= \int \frac{cos \theta}{\sqrt{cos^{2}\theta}} d\theta[/itex]

    [itex]= \int \frac{cos \theta}{{cos\theta}} d\theta[/itex]

    [itex]= \int 1\ d\theta[/itex]

    [itex]= \theta[/itex]

    [itex]x = sin \theta[/itex]

    [itex]\theta = arcsin x[/itex]


    second attempt (the wrong answer I believe)

    [itex]let\ x = cos \theta[/itex]

    [itex]dx = -sin \theta\ d\theta[/itex]

    [itex]\int \frac{-sin \theta}{\sqrt{1-cos^{2}\theta}} d\theta[/itex]

    [itex]= \int \frac{-sin \theta}{\sqrt{sin^{2}\theta}} d\theta[/itex]

    [itex]= \int \frac{-sin \theta}{{sin\theta}} d\theta[/itex]

    [itex]= \int -1\ d\theta[/itex]

    [itex]= -\theta[/itex]

    [itex]x = cos \theta[/itex]

    [itex]\theta = -arccos x[/itex]
     
  2. jcsd
  3. Dec 12, 2013 #2
    Its simple...we know that d/dx arc sinx = (1-x^2)^-1/2
    d/dx arc cos x = -(1-x^2)^-1/2 (notice the minus sign)
    as integral is antiderivative of a function you are getting the same values
    yes integral arc sinx=(minus)integral arc cos x..Hope your doubt is clarified
     
  4. Dec 12, 2013 #3

    ShayanJ

    User Avatar
    Gold Member

    In fact [itex] \int \frac{1}{\sqrt{1-x^2}}dx=\sin^{-1}{x}+C_1 \ and \ \int \frac{1}{\sqrt{1-x^2}}dx=-\cos^{-1}{x}+C_2 [/itex]
    If you take a look at the plots of the functions [itex] \sin^{-1}{x} [/itex] and [itex] \cos^{-1}{x} [/itex],you can see that you can get one of them by multiplying the other by a minus sign and then adding a constant(or vice versa!)...and you can see that you have the negative sign and also the constant in the results of the integral!
     
  5. Dec 12, 2013 #4
    Thanks for clearing that up for me.
     
  6. Dec 12, 2013 #5

    Mark44

    Staff: Mentor

    No one has mentioned this. There's a trig identity that can shed some light here.

    ##sin^{-1}(x) + cos^{-1}(x) = \pi/2##
    The two functions differ by a constant. If an indefinite integral produces two different antiderivatives, those functions can differ by at most a constant.
     
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