# Quick Problem with an integral

1. Oct 15, 2012

### mattyc33

I am just having problems factoring the constants out of the integral:

∫Fsinwt dt

where F and w are constants.

I can obviously take F right out of the integral but I forget how to take the w out of the sin, and I can't seem to find anywhere on the internet that will tell me how to do so.

If anyone could quickly help me that would be great. Thanks.

EDIT: If anyone could explain to me how to enter an integral like this into wolfram alpha so I can have steps that would be just as good.

2. Oct 15, 2012

### Gloyn

What is the final porpose of doing that?
You could use the half-angle formula:

sin(2x)=2sinxcosx

for particular 'w', but I don't think that there is a general (resonable) formula for any w. The only thing that comes to my mind is using de Moivre's theorm:

(cosx+isinx)^w=(coswx+isinwx)=(cosx+isinx)^w, and now you can multiply out the bracket and take the imaginary part what wil leave u with:

sinwx=(powers of sinx)

3. Oct 15, 2012

### zapz

You don't need to factor out the w. You really just have [tex]F\intsinwtdt[\tex]. You just need to integrate sinwt. So you'll get -coswt, and accounting for the w and F gives (-Fcoswt)/w. For wolfram alpha just type in integrate and then the function

4. Oct 15, 2012

### Gloyn

Yeah, but I think that he has some hidden purpose in doing that trick :P It's pretty simple to integrate, so propably he needs it in that form.

5. Oct 15, 2012

### haruspex

The way I find easiest is 'everywhere I see a t, replace it with t/w'. Of course, that includes in the bounds of the integral.

6. Oct 15, 2012

### mathman

u = wt, so dt = du/w. The integral becomes (F/w)∫sinudu = -(F/w)coswt + C

Not acquainted with Wolfram alpha.

7. Oct 15, 2012

### dextercioby

In other words, even if the constants are abstractly depicted through letters, you must still be able to apply the elementary methods like part integration or substitution.

8. Oct 15, 2012

### Gloyn

Well, I still think that the problem was not to simply evaluate the integral, which is easy, but to evaluate it by a given method (why else struggle with dragging that constant out of the sin?).