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    I just derived the integral of atan x in two different ways [by parts and by considering it as (log(1+ix)-log(1-ix))/(2i)], and after a bit of fiddling about got the same result, but there was a +1 in there when I did it using logarithms. I checked it on Wolfram, using atan x itself and its equivalent, and it came up with the same.
    I know it doesn't really matter given that you're either going to add a constant anyway or it'll dissappear upon applying limits, but it bugged me. Can anyone explain why this is, or is it just a case of "that's just the way it is; don't worry about it"?
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    Nothing is ever a case of "that's just the way it is". Unfortunately, I don't know what you're talking about. :p: When you integrate, try adding an arbitrary constant, and then set x equal to some specific value to work out that constant. Sounds like it might fix your problem.
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    the a + 1 is just part of the arbitary constant.
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    For clarification, this is what I did:

    By parts

    \displaystyle\int\arctan x\mathrm d x = \displaystyle\int\arctan x\cdot 1\mathrm d x \\= x\arctan x - \displaystyle\int\frac x{x^2+1}\\  \mathrm{\qquad let\ } u=x^2+1\Rightarrow\mathrm d u=2x\mathrm{\ d} x\Rightarrow\mathrm d x=\frac{\mathrm d u}{2x}\\=x\arctan x - \frac 12\displaystyle\int\frac 1u\\=x\arctan x - \frac 12 \log(x^2+1) +C


    Using logs

    But you could alternatively say that:
    \displaystyle{\tan x = y=\frac{\sin x}{\cos x} = -i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}=-i\frac{e^{2ix}-1}{e^{2ix}+1}}\\ \Rightarrow iy(e^{2ix}+1)=e^{2ix}-1\\ \Rightarrow e^{2ix}(iy-1)=-(iy+1)\\ \Rightarrow e^{2ix}=\frac{1+iy}{1-iy}\\ \Rightarrow x=\frac{-i}2 (\log(1+iy)-\log(1-iy))

    So
    \arctan t = \frac{-i}2 (\log(1+it)-\log(1-it))\\ \Rightarrow \displaystyle\int\arctan t \mathrm{\ d}t = \frac{-i}2 \displaystyle\int(\log(1+it)-\log(1-it)) \mathrm{\ d}t\\ \\

\mathrm{\qquad} \begin{array}{lcl} \mathrm{let\ } u=1+it & & \mathrm{let\ } v=1-it

\\\mathrm {d} u=i\mathrm {\ d} t & & \mathrm {d} v=-i\mathrm {\ d} t 

\\ \mathrm {d} t=-i\mathrm {\ d} u & & \mathrm d t= i\mathrm {\ d} v \end{array} \\ \\

=\frac{-i}2\left[ -i\displaystyle\int \log u\mathrm{\ d} u -i\displaystyle\int \log v\mathrm{\ d} v  \right]\\

\frac {-i}2 \left[ -i(1-it)(\log(1-it)-1)-i(1+it)(\log(1+it)-1) \right] +C\\

=-\frac 12 \left[ (1-it) \log(1-it) -1+it +(1+it) \log(1+it) -1-it\right] +C\\

-\frac 12 \left[ \log(1-it) +\log(1+it) +it(log(1+it)-\log(1-it)) \right] +1+C\\=

\frac{-it}{2}(log(1+it)-\log(1-it)) - \frac 12 \log (1+x^2) +1+C\\=

t\arctan t -\frac 12 \log (1+x^2) +1+C



    The two methods give the same result, except for the strange 1.
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    (Original post by matt2k8)
    the a + 1 is just part of the arbitary constant.
    I know, but why does it appear with one method and not the other?
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    lol i thought this thread was about satan
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    (Original post by cconstant)
    lol i thought this thread was about satan
    :sadnod:
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    (Original post by meatball893)
    I know, but why does it appear with one method and not the other?
    If C is an arbitrary constant, then C + 1 is also an arbitrary constant, so it's still an arbitrary constant; the 1 doesn't matter (it emerged by using a different method). You could just as well let D = C + 1, where D is an arbitrary constant, and then the two results are equivalent.
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    I thought it had something to do with satan too.
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    I know it doesn't matter but it just seems strange that there's a +1 in one and not in the other. Oh well.
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    A discussion about satan? In the maths forum? :rolleyes:
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    This thread is now about Satan.
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    (Original post by meatball893)
    For clarification, this is what I did:

    By parts

    \displaystyle\int\arctan x\mathrm d x = \displaystyle\int\arctan x\cdot 1\mathrm d x \\= x\arctan x - \displaystyle\int\frac x{x^2+1}\\  \mathrm{\qquad let\ } u=x^2+1\Rightarrow\mathrm d u=2x\mathrm{\ d} x\Rightarrow\mathrm d x=\frac{\mathrm d u}{2x}\\=x\arctan x - \frac 12\displaystyle\int\frac 1u\\=x\arctan x - \frac 12 \log(x^2+1) +C


    Using logs

    But you could alternatively say that:
    \displaystyle{\tan x = y=\frac{\sin x}{\cos x} = -i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}=-i\frac{e^{2ix}-1}{e^{2ix}+1}}\\ \Rightarrow iy(e^{2ix}+1)=e^{2ix}-1\\ \Rightarrow e^{2ix}(iy-1)=-(iy+1)\\ \Rightarrow e^{2ix}=\frac{1+iy}{1-iy}\\ \Rightarrow x=\frac{-i}2 (\log(1+iy)-\log(1-iy))

    So
    \arctan t = \frac{-i}2 (\log(1+it)-\log(1-it))\\ \Rightarrow \displaystyle\int\arctan t \mathrm{\ d}t = \frac{-i}2 \displaystyle\int(\log(1+it)-\log(1-it)) \mathrm{\ d}t\\ \\

\mathrm{\qquad} \begin{array}{lcl} \mathrm{let\ } u=1+it & & \mathrm{let\ } v=1-it

\\\mathrm {d} u=i\mathrm {\ d} t & & \mathrm {d} v=-i\mathrm {\ d} t 

\\ \mathrm {d} t=-i\mathrm {\ d} u & & \mathrm d t= i\mathrm {\ d} v \end{array} \\ \\

=\frac{-i}2\left[ -i\displaystyle\int \log u\mathrm{\ d} u -i\displaystyle\int \log v\mathrm{\ d} v  \right]\\

\frac {-i}2 \left[ -i(1-it)(\log(1-it)-1)-i(1+it)(\log(1+it)-1) \right] +C\\

=-\frac 12 \left[ (1-it) \log(1-it) -1+it +(1+it) \log(1+it) -1-it\right] +C\\

-\frac 12 \left[ \log(1-it) +\log(1+it) +it(log(1+it)-\log(1-it)) \right] +1+C\\=

\frac{-it}{2}(log(1+it)-\log(1-it)) - \frac 12 \log (1+x^2) +1+C\\=

t\arctan t -\frac 12 \log (1+x^2) +1+C



    The two methods give the same result, except for the strange 1.
    :facepalm: The constant C is just a number. So the C you achieved by substitution is +1 greater than the C you achieved by logs. So in fact the C+1 in your answer by logs can just go to C as it is again just a number.
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    I know it's just a constant, but when I did the two methods I fully expected them to come out with exactly the same result, and so when I saw a 1 it threw me. Don't know why.
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    (Original post by meatball893)
    I know it's just a constant, but when I did the two methods I fully expected them to come out with exactly the same result, and so when I saw a 1 it threw me. Don't know why.
    There is no need for it to throw you. C is just a number and all it means that the C in one method is not the same as the C in the other method(differing by 1) Thats all. I had to explain that to my maths teacher before, as the same occurred integrating something else, and caused quite a bit of confusion
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    It threw me too when I saw it. It appears more when integrating by parts because you get more than one term automatically.

    For another example where it crops up, which should convince you that +C really can change absolutely as needed, try integrating ∫1/x dx by parts as x and x^-2.

    Spoiler:
    Show
    ∫1/x dx = ∫1/x dx +1 = ∫1/x dx +2 = ...

    (=lnx +C)
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    Yes they are the same thing but he's asking does anyone know WHY that happens ?
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    (Original post by spex)
    Yes they are the same thing but he's asking does anyone know WHY that happens ?
    By parts, you put off adding the constant until right at the end. With the other method, you do the actual integration more or less straight away, and if you follow it you'll find that the 1 comes from writing \tan x as -i \dfrac{e^{ix} - e^{-ix}}{e^{ix}+e^{-ix}} and the subsequent substitutions.

    Take this as another example:
    \displaystyle \newline

\dfrac{d}{dx}\sin^2 x = 2\sin x \cos x\newline

\Rightarrow \int 2\sin x \cos x\ \mbox{d}x = \boxed{\sin^2 x + C}\newline

\dfrac{d}{dx}\cos^2 x = -2\sin x \cos x\newline

 \Rightarrow \int 2\sin x \cos x\ \mbox{d}x = -\cos^2 x + C\newline

 = -(1-\sin^2 x) + C\newline

 = \boxed{\sin^2 - 1 + C}

    The differences occur in the method you use to get to the answer. It's not strange, and I can't understand why it would throw you.
 
 
 
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