How to Factor Constants Out of Trigonometric Integrals?

[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.

 

[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 theorem:

(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)

 

[zapz]

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

 

[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.

 

[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.

 

[mathman]

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.

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

Not acquainted with Wolfram alpha.

 

[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.

 

[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?).