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    Hey guys,

    I am stuck once again.

    i) Let I be the integral in this case... I cos^-1(x) with x=cos u

    My workings....

    if x=cosU
    dx= -sinU

    Therefore I cos^-1(cos(U)) * -sin U = I -Usin U

    Now using integration by parts, You get UcosU - I Cos U

    Now if cosU =x, then U = cos^-1(x)
    Then, Cos^-1(x) * Cos(Cos^-1(x)) - Sin(Cos^-1(x))

    Then you get XCos^-1(x) -Sin(Cos^-1(x))

    The book says the answe is XCOS^-1(X) - SQRT* 1-X^2 +K

    How the hell did they get that?

    I am so confused
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    They're the same answer, since \sin (\cos^{-1} x) = \sqrt{1-x^2}, which you can see by writing \sin \theta = \sqrt{1-\cos^2 \theta}; and here \theta = \cos^{-1} x.
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    http://en.wikipedia.org/wiki/Inverse...tric_functions

    Have a look at the relations in Section 2.
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    (Original post by nuodai)
    They're the same answer, since \sin (\cos^{-1} x) = \sqrt{1-x^2}, which you can see by writing \sin \theta = \sqrt{1-\cos^2 \theta}; and here \theta = \cos^{-1} x.
    ahh right I see thanks!

    Just one quick question, without starting a new thread...

    If U= e^x
    then what would U be if it is e^-x?

    Would it be 1/u?
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    (Original post by J DOT A)
    ahh right I see thanks!

    Just one quick question, without starting a new thread...

    If U= e^x
    then what would U be if it is e^-x?

    Would it be 1/u?
    If I understood what you said correctly, yes, e^{x} = u \implies e^{-x} = \frac{1}{u}.
 
 
 
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