$\displaystyle \int \frac{2}{(4u^2 + 1)} du $

$\displaystyle s = 2u$

$\displaystyle = \int \frac{2}{(s^2 + 1)} ds $

$\displaystyle = 2 \int \frac{1}{(s^2 + 1)} ds $

$\displaystyle = tan^{-1} (2u)$

where does the 2 go?

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- Jun 12th 2013, 03:36 AMalyosha2Where does the 2 go?
$\displaystyle \int \frac{2}{(4u^2 + 1)} du $

$\displaystyle s = 2u$

$\displaystyle = \int \frac{2}{(s^2 + 1)} ds $

$\displaystyle = 2 \int \frac{1}{(s^2 + 1)} ds $

$\displaystyle = tan^{-1} (2u)$

where does the 2 go? - Jun 12th 2013, 03:42 AMchiroRe: Where does the 2 go?
Hey alyosha2.

You've screwed up your integration variables but the idea is that ds = 2du so you replace 2*du with ds and the 2 disappears along with the du after a substitution. - Jun 12th 2013, 03:49 AMalyosha2Re: Where does the 2 go?
what do you mean by ds = 2du exactly?

- Jun 12th 2013, 03:54 AMchiroRe: Where does the 2 go?
Your substitution is s = 2u. Differentiating gives ds/du = 2 and for integral substitution this means ds = 2du.

- Jun 12th 2013, 04:10 AMalyosha2Re: Where does the 2 go?
But ds and dx are just markers reflecting what variable we using in the intergrand. So if the variable we switch it to is twice that of the one we switched it from, I don't see how that effects anything in the integrand.

- Jun 12th 2013, 04:23 AMchiroRe: Where does the 2 go?
(2u)^2 = 4u^2 = s^2 which means the 4 disappears after the substitution.

- Jun 12th 2013, 04:30 AMalyosha2Re: Where does the 2 go?
sorry I edited my post. Didn't mean to ask that. See above.

- Jun 12th 2013, 04:37 AMalyosha2Re: Where does the 2 go?
if s = 2u then any 2u becomes an s. But I can't see the how s = 2u makes a 2 disappear.

- Jun 12th 2013, 04:44 AMalyosha2Re: Where does the 2 go?
So what you seem to be saying here is we take out the 2 from the integral and place in the marker notation!?!? Now I'm really confused. Is this supposed to be some kind of short hand? It doesn't make any sense other wise. The du isn't a term of the expression, it's just a marker showing us what letter we are using for the variable. How can we start treating it as a term or factor? I don't understand.

- Jun 12th 2013, 05:26 AMalyosha2Re: Where does the 2 go?
OK, so the way I'm thinking about it now is that when you do the substitution you're changing the integrand into an expression that operates on a variable that is twice that of the original variable. So to keep the integrand equal to the value of the original expression you need to keep the balance somehow. When I put the above integrand into wolfram, but with a 1 in place of the 2, it pulls out a 1/2 to do this. So there is a link here obviously between the s being twice that of u, but I can't seem to quite grasp it. Does simply dividing the whole integrand by 2 keep the integrand equal to the first variable, it obviously does, but I can't quite grasp why is should. Can anyone explain it to me?

- Jun 12th 2013, 06:12 AMSorobanRe: Where does the 2 go?
Hello, alyosha2!

All this discussion is going nowhere . . .

Quote:

$\displaystyle \displaystyle\int \frac{2}{(4u^2 + 1)}\, du $

$\displaystyle \text{Let }s \,=\,2u \quad\Rightarrow\quad ds \,=\,2\;\!du \quad\Rightarrow\quad du \,=\,\tfrac{1}{2}\;\!ds$

$\displaystyle \displaystyle\text{Substitute: }\:\int\frac{2}{s^2+1}\left(\tfrac{1}{2}\;\!ds \right) \;=\;\int\frac{ds}{s^2+1} $

Got it?