Very basic antidifferentiation problem

**Wolvenmoon** · March 30th 2010, 03:42 PM

I still don't know the math symbols, so please bear with me!

I have the big S thing over sin(2X)*DX, its antidifferentiation is -(cos(2X))/2 + C

I have completely lost where that over 2 comes from. Wolfram alpha is not helpful. I realize that somewhere we pull a 1/2 out in front of everything, but I don't know where.

Thanks, I hope I'm making SOME sense.

**pickslides** · March 30th 2010, 03:49 PM

Remember that

$\int \sin nx ~dx = -\frac{1}{n}\cos nx +C$

Therefore

$\int \sin 2x ~dx = -\frac{1}{2}\cos 2x +C$

**skeeter** · March 30th 2010, 03:53 PM

Originally Posted by Wolvenmoon

I still don't know the math symbols, so please bear with me!

I have the big S thing over sin(2X)*DX, its antidifferentiation is -(cos(2X))/2 + C

I have completely lost where that over 2 comes from. Wolfram alpha is not helpful. I realize that somewhere we pull a 1/2 out in front of everything, but I don't know where.

Thanks, I hope I'm making SOME sense.

formally ...

$\int \sin(2x) \, dx<br />$

$u = 2x$

$du = 2 \, dx$

$\frac{1}{2} \int \sin(2x) \cdot 2 \, dx$

substitute ...

$\frac{1}{2} \int \sin(u) \, du<br />$

integrate ...

$-\frac{1}{2} \cos(u) + C<br />$

back-substitute ...

$-\frac{1}{2} \cos(2x) + C$

informally ...

$\int \sin(kx) \, dx$

the antiderivative of $\sin(kx)$ has to be a form of $-\cos(kx)$ , but taking the derivative of $-\cos(kx)$ yields $\sin(kx) \cdot k$ because of the chain rule ... now, what do you think will get rid of the constant factor $k$ ?

**Wolvenmoon** · March 30th 2010, 08:56 PM

Okay Skeeter, I went back over this and there are two steps in particular that have me. When the 1/2 shows up in front of everything, how did it get there and why?

My instructor has taken a completely different approach to this than my book did, hers is an easier way but I think I've managed to drop something somewhere.

**pickslides** · March 31st 2010, 01:29 AM

Originally Posted by Wolvenmoon

When the 1/2 shows up in front of everything, how did it get there and why?

Find $\frac{d}{dx}\left(-\frac{1}{n}\cos nx +C\right)$

That should make things clearer.

**HallsofIvy** · March 31st 2010, 01:35 AM

When you are substituting into $\int sin(2x) dx$ , you have to substitute for every "x", even the one is "dx".

As skeeter said, he let u= 2x. Differentiating both sides of that, du= 2 dx.

There are two ways of thinking about that but they give the same result:

1) Multiply and divide the integral by 2-
$\int sin(2x)\left(\frac{1}{2}\right)(2x) dx= \frac{1}{2}\int sin(2x) 2dx$ so your integral becomes $\frac{1}{2}\int sin(u)du$ .

2) Divide both sides of du= 2 dx by 2 to get $\frac{1}{2} du= dx$ and replace dx by that-
$\int sin(u) \frac{1}{2} du= \frac{1}{2}\int sin(u) du$ .

Notice that it is the fact that " $\frac{1}{2}$ " is a constant that allows us to move it in or out of the integral at will.

**Wolvenmoon** · March 31st 2010, 10:39 AM

So essentially that's 2/2 or a form of 1 that we're adding into the picture to force du = 2dx, breaking up into 2/1 and 1/2, and moving the 1/2 over while we work?

Okay. Think I've cleared the brick wall here. Thanks!

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