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∫ arctan x

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"?

Reply 1

generalebriety

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.

Reply 2

username219321

the a + 1 is just part of the arbitary constant.

Reply 3

meatball893

For clarification, this is what I did:

By parts

Using logs

The two methods give the same result, except for the strange 1. :s-smilie:

Reply 4

meatball893

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?

Reply 5

cconstant

lol i thought this thread was about satan :s-smilie:

Reply 6

ashy

cconstant

lol i thought this thread was about satan :s-smilie:

Reply 7

nuodai

meatball893

I know, but why does it appear with one method and not the other?

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.

Reply 8

Fiasco

I thought it had something to do with satan too.

Reply 9

meatball893

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.

Reply 10

meatball893

A discussion about satan? In the maths forum? :rolleyes:

Reply 11

Phugoid

This thread is now about Satan.

Reply 12

It could be lupus

meatball893

For clarification, this is what I did:

By parts

Using logs

The two methods give the same result, except for the strange 1. :s-smilie:

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.

Reply 13

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.

Reply 14

It could be lupus

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

Reply 15

Meeton

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

Reply 16

spex

Yes they are the same thing but he's asking does anyone know WHY that happens ?

Reply 17

nuodai

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

-i \dfrac{e^{ix} - e^{-ix}}{e^{ix}+e^{-ix}}

and the subsequent substitutions.

Take this as another example:

Unparseable latex formula:

\displaystyle \newline[br]\dfrac{d}{dx}\sin^2 x = 2\sin x \cos x\newline[br]\Rightarrow \int 2\sin x \cos x\ \mbox{d}x = \boxed{\sin^2 x + C}\newline[br]\dfrac{d}{dx}\cos^2 x = -2\sin x \cos x\newline[br] \Rightarrow \int 2\sin x \cos x\ \mbox{d}x = -\cos^2 x + C\newline[br] = -(1-\sin^2 x) + C\newline[br] = \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.