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?

Printable View

- July 19th 2009, 02:55 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? - July 19th 2009, 02:58 PMskeeter
- July 19th 2009, 03:08 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? - July 19th 2009, 03:10 PMskeeter
no

- July 19th 2009, 03:14 PMBrazuca
how come the u^2 isn't effected by the 1/2?

- July 19th 2009, 03:35 PMskeeter

note that the on the outside of the integral compensates for the 2 inserted in the integrand

see why now? - July 19th 2009, 03:47 PMBrazuca
Oh now I understand. So by having the 1/2 on the outside it causes the 2 to appear on the inside.

- July 19th 2009, 03:53 PMskeeter
- July 19th 2009, 04:18 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? - July 19th 2009, 04:21 PMskeeter
- July 19th 2009, 04:31 PMPlato