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    integrate (cos^-1(x))((1-x^2)^(1/2))

    i try to solve this for the entire day nut still not succeed .please help me out.
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    Try the substituion x=\cos{\theta}

    Remeber \cos^{-1}{(\cos{\theta})} = \theta

    I don't know if it any easier to solve but it looks a little nicer.

    Edit: It should come out with integration by parts after this.
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    (Original post by The Muon)
    Remeber \cos^{-1}{(\cos{\theta})} = \theta
    For -\pi< \theta <  0, \cos^{-1}{(\cos{\theta})} = - \theta.

    EDIT: Although I think it is true that \arccos(\cos(\theta)) \cdot \sqrt{\sin^2(\theta)} is always equal to \theta \sin(\theta), which makes the integral easier to deal with.

    2nd EDIT: Ignore the above, the substitution works fine with 0 \leq \theta \leq \pi
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    (Original post by Mark13)
    For -\pi< \theta <  0, \cos^{-1}{(\cos{\theta})} = - \theta.

    EDIT: Although I think it is true that \arccos(\cos(\theta)) \cdot \sqrt{\sin^2(\theta)} is always equal to \theta \sin(\theta), which makes the integral easier to deal with.

    2nd EDIT: Ignore the above, the substitution works fine with 0 \leq \theta \leq \pi
    no, don't be a retard. f^-1(f(x))=x.
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    I'm always on the top

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    Love that songz
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    WTF?? So glad Im not doing maths beyond GCSE, i struggled to get an A in gcse lol
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    (Original post by adamdivine)
    I'm always on the top

    Tonight I'm on the bottom

    We trading places


    Love that songz
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    (Original post by Totally Tom)
    no, don't be a retard. f^-1(f(x))=x.
    :rolleyes:

    What is \arccos\frac{1}{2}? \frac{\pi}{3}

    What is \cos-\frac{\pi}{3}? \frac{1}{2}

    So \arccos(\cos (-\frac{\pi}{3}))=\frac{\pi}{3}.

    The range of the function arccos is limited to 0 \leq \arccos(x) \leq \pi.
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    (Original post by Mark13)
    :rolleyes:

    What is \arccos\frac{1}{2}? \frac{\pi}{3}

    What is \cos-\frac{\pi}{3}? \frac{1}{2}

    So \arccos(\cos (-\frac{\pi}{3}))=\frac{\pi}{3}.

    The range of the function arccos is limited to 0 \leq \arccos(x) \leq \pi.
    arccos(1/2) also equals -pi/3.
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    2 + 3 = 5. These words will reverberate through the generation
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    (Original post by Totally Tom)
    arccos(1/2) also equals -pi/3.
    No it doesn't
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    (Original post by adamdivine)
    2 + 3 = 5. These words will reverberate through the generation
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    Why are you doing maths?
    It's holidayss!
    I am never looking at integration ever again! Hate it!!
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    (Original post by Mark13)
    No it doesn't
    i'm not going to argue with you about this.
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    (Original post by Mark13)
    No it doesn't
    see here
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    (Original post by Totally Tom)
    i'm not going to argue with you about this.
    There's nothing to argue about; if arccos was a one-to-many mapping, it would be impossible to integrate.
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    (Original post by bebeh)
    integrate (cos^-1(x))((1-x^2)^(1/2))

    i try to solve this for the entire day nut still not succeed .please help me out.
    What do you get if you differentiate arcos(x)?
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    (Original post by Mark13)
    There's nothing to argue about; if arccos was a one-to-many mapping, it would be impossible to integrate.
    this is simply convention, arccos is multi valued, just like sqrt(x) is multivalued, it's just we don't count most of them.
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    (Original post by Totally Tom)
    this is simply convention, arccos is multi valued, just like sqrt(x) is multivalued, it's just we don't count most of them.
    Whenever we talk about arccos(x) in the context of calculus, it is a one-to-one mapping.Similarly, when \sqrt{x} appears in an integrand, it is defined as the postive square root of x.
 
 
 
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