Laplace Transform of e-t sin t: How to Derive the Denominator?

[trojansc82]

Homework Statement

Laplace Transform of e^-t sin t

Homework Equations

The Attempt at a Solution

I have the solution, but I am unable to figure out how the denominator becomes 1/[(s + 1)² + 1]

 

[rock.freak667]

The presence of the e^at would case the shift from 's' to 's-a'. This is why it is called the shift theorem, it's mainly used in the inverse laplace transform.

So you know that L{sint} = 1/(s²+1)

and following shift theorem L(e^atsint) = 1/[(s-a)²+1].

You can derive it too using the integral formula.

 

[trojansc82]

The presence of the e^at would case the shift from 's' to 's-a'. This is why it is called the shift theorem, it's mainly used in the inverse laplace transform.

So you know that L{sint} = 1/(s²+1)

and following shift theorem L(e^atsint) = 1/[(s-a)²+1].

You can derive it too using the integral formula.

I am unable to derive it from the integral formula. I need to see the steps. I'm fairly certain I've been able to integrate it correctly, but I keep getting a repetitive e^-t sin t or e^-t cos t when I integrate.

 

[Bohrok]

I am unable to derive it from the integral formula. I need to see the steps. I'm fairly certain I've been able to integrate it correctly, but I keep getting a repetitive e^-t sin t or e^-t cos t when I integrate.

So you're probably ending up with something like ∫e^-t sin t dt on both sides of the equation; just move one to the other side and combine them as like terms. There's a good example of a similar problem on Wikipedia with ∫e^x cos x dx:
http://en.wikipedia.org/wiki/Integration_by_parts#Integrals_with_powers_of_x_or_ex