http://img24.imageshack.us/img24/7753/beh.png

I don't understand the circled part. Since U uses up the 2x in front of the cos2x how come a DU is still present. Also why is the 1/2 there?

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- Jul 19th 2009, 01:55 PMBrazucaAnti-Derivative using u and du.
http://img24.imageshack.us/img24/7753/beh.png

I don't understand the circled part. Since U uses up the 2x in front of the cos2x how come a DU is still present. Also why is the 1/2 there? - Jul 19th 2009, 01:58 PMskeeter
- Jul 19th 2009, 02:08 PMBrazuca
Then what about the u^2?

why is it not effected by the 1/2?

shouldn't Sin(squared)2x turn into [(squared)2x]/2? - Jul 19th 2009, 02:10 PMskeeter
no

- Jul 19th 2009, 02:14 PMBrazuca
how come the u^2 isn't effected by the 1/2?

- Jul 19th 2009, 02:35 PMskeeter
$\displaystyle \int \textcolor{red}{\sin^2(2x)} \textcolor{blue}{\cos(2x) \, dx}$

$\displaystyle \textcolor{red}{u = \sin(2x)}$

$\displaystyle \textcolor{blue}{du = 2\cos(2x) \, dx}$

$\displaystyle \textcolor{blue}{\frac{1}{2}} \int \textcolor{red}{\sin^2(2x)} \textcolor{blue}{2\cos(2x) \, dx}$

note that the $\displaystyle \frac{1}{2}$ on the outside of the integral compensates for the 2 inserted in the integrand

$\displaystyle \textcolor{blue}{\frac{1}{2}} \int \textcolor{red}{u^2} \textcolor{blue}{du}$

see why now? - Jul 19th 2009, 02:47 PMBrazuca
Oh now I understand. So by having the 1/2 on the outside it causes the 2 to appear on the inside.

- Jul 19th 2009, 02:53 PMskeeter
- Jul 19th 2009, 03:18 PMBrazuca
Ok now I think I get it.

Because you put the 2 inside the integral you need to compensate by putting the 1/2 on the outside.

So if I had needed to put a 5 inside I would put a 1/5 outside right? - Jul 19th 2009, 03:21 PMskeeter
- Jul 19th 2009, 03:31 PMPlato