Finding Laplace Transforms Sin(bt)

[Lee49645]

Homework Statement

Simply find L{Sin(bt)}

Homework Equations

My Professor gave us a hint "Use Integration by Parts TWICE"

The Attempt at a Solution

so i start with the usual limit as N -> infinity the integral of e^(-st)sin(bt)dt

Integrating by parts i get ( u = sin(bt) du = bcos(bt) dv = e^(-st) v = (-1/s)e^(-st))

sin(bt)(-1/s)e^(-st) - integral((-1/s)e^(-st) bcos(bt))

cleaning up the problem the integral becomes e^(-st)cos(bt)dt.

This is an endless cycle because after integrating i ended up with an extra uv set and integral e^(-st)sin(bt)dt with a b^2 outside.

it looks nothing like the known answer b/(s^s + b^s)

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Gregg]

If you are asked to do the integration by parts twice, what do you notice the second time you do it? The last term is an integral of something very similar to the LHS. Take it away from both sides.

[tex]\int e^{ux} \cos x dx= e^{ux} \sin x - \int ue^{ux} \sin x dx [/tex]

[tex]\int e^{ux} \cos x dx= e^{ux} \sin x +ue^{ux} \cos x - \int u^2e^{ux} \cos x dx][/tex]

[tex]\int (1+u^2)e^{ux} \cos x dx = e^{ux} \sin x + ue^{ux} \cos x[/tex]

[tex]\int e^{ux} \cos x dx = \frac{e^{ux} \sin x + ue^{ux} \cos x}{(1+u^2)}[/tex]

 

[Char. Limit]

If you are asked to do the integration by parts twice, what do you notice the second time you do it? The last term is an integral of something very similar to the LHS. Take it away from both sides.

[tex]\int e^{ux} \cos x dx= e^{ux} \sin x - \int ue^{ux} \sin x dx [/tex]

[tex]\int e^{ux} \cos x dx= e^{ux} \sin x +ue^{ux} \cos x - \int u^2e^{ux} \cos x dx][/tex]

[tex]\int (1+u^2)e^{ux} \cos x dx = e^{ux} \sin x + ue^{ux} \cos x[/tex]

[tex]\int e^{ux} \cos x dx = \frac{e^{ux} \sin x + ue^{ux} \cos x}{(1+u^2)}[/tex]

Considering that both u and u^2 are constants, why didn't you just pull them out of the integral? It makes it easier in my opinion...

 

[Gregg]

Yeah u is a constant, but the point of what I've done is made it so there is no integral that carries on forever on the RHS because I got one that is similar enough to the LHS so that I could subtract it from both sides and have an expression for the integral that's useful. This is what I think you are supposed to do when you are asked to integrate by parts twice.

 

[Gregg]

For you, u=-s and cos x = sin (bt)

 

[Char. Limit]

Heh, I usually just evaluated the bounds as I went, instead of waiting till the end. I'd show you what I mean, except for two things:

1. I don't want to give the answer away...
2. You likely already get what I mean.

 

[Gregg]

There are no bounds, it is an indefinite integral. I think you mean you would not wait til the end to substitute the values as I have described in my demonstration. I would not wait until the end either, I gave the example and the way which the OP can do this problem for themselves from the beginning and I didnt intend for them to substitute the variables in in the final stage. If you did this, in general, you will not get the correct answer.

 

[HallsofIvy]

On the contrary, there are bounds. The Laplace transform integrates from 0 to infinity.

 

[Gregg]

I was just referring to the integration of e^x sin x or e^x cos x, and making the distinction between the bounds of the integral in which the example has none of, and the substition that the OP should make in order to do this for themselves which cannot be done at the end point.

 

[Lee49645]

If you are asked to do the integration by parts twice, what do you notice the second time you do it? The last term is an integral of something very similar to the LHS. Take it away from both sides.

[tex]\int e^{ux} \cos x dx= e^{ux} \sin x - \int ue^{ux} \sin x dx [/tex]

[tex]\int e^{ux} \cos x dx= e^{ux} \sin x +ue^{ux} \cos x - \int u^2e^{ux} \cos x dx][/tex]

[tex]\int (1+u^2)e^{ux} \cos x dx = e^{ux} \sin x + ue^{ux} \cos x[/tex]

[tex]\int e^{ux} \cos x dx = \frac{e^{ux} \sin x + ue^{ux} \cos x}{(1+u^2)}[/tex]

wow you're right, after looking up the formula for integration by parts I forgot that there was also an integral on the LHS. though my answer doesn't match yours, I actually got the answer i needed.



Instead of subtracting, I brought the matching integrals together on the LHS and factored them out leaving an [(s^2 + b^2) / s^2] (integral[blah])

then evaluating the remaining equations on the RHS from 0 to infinity, I was left with b/s^2

then in my last step I just multiplied both sides by the reciprocal leaving my final answer of

integral(e^(-st)sin(bt)dt) = (b/s^2) (s^2/(s^2 + b^2)) => b/(s^2 + b^2)

 

[Gregg]

Great, you shouldn't get the same answer as me anyway I integrated cosine.