Anti-Derivative using u and du.

**Brazuca** · July 19th 2009, 01:55 PM

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?

**skeeter** · July 19th 2009, 01:58 PM

Originally Posted by Brazuca

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?

note that $\frac{du}{2} = \cos(2x) \, dx$

**Brazuca** · July 19th 2009, 02:08 PM

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?

**skeeter** · July 19th 2009, 02:10 PM

no

**Brazuca** · July 19th 2009, 02:14 PM

how come the u^2 isn't effected by the 1/2?

**skeeter** · July 19th 2009, 02:35 PM

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

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

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

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

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

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

see why now?

**Brazuca** · July 19th 2009, 02:47 PM

Oh now I understand. So by having the 1/2 on the outside it causes the 2 to appear on the inside.

**skeeter** · July 19th 2009, 02:53 PM

Originally Posted by Brazuca

Oh now I understand. So by having the 1/2 on the outside it causes the 2 to appear on the inside.

other way around ... you need the constant 2 in the integrand for du, the 1/2 on the outside compensates because (1/2)*2 = 1

**Brazuca** · July 19th 2009, 03:18 PM

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?

**skeeter** · July 19th 2009, 03:21 PM

Originally Posted by Brazuca

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?

that is correct. in effect, it's like multiplying the integral by 1 ... no change.

**Plato** · July 19th 2009, 03:31 PM

Originally Posted by Brazuca

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?

I have a suggestion; simply differentiate the following paying very close attention to the process.
$\frac{1}{6}\sin^3(2x)$

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