Find Laplace transform for the following two functions eat s

Find Laplace transform for the following two functions e^-at sin (wt) e^-at cos (wt)

Solution

a)
Let I = e^(-st) * e^(-at) sin(wt) dt.
(We will deal with the bounds later.)

Rewrite this as I = e^(-(s+a)t) sin(wt) dt.

Now, use integration by parts:
Let u = e^(-(s+a)t), dv = sin(wt) dt
and du = -(s+a) e^(-(s+a)t) dt, v = (-1/w) cos(wt).

So, I = (-1/w) e^(-(s+a)t) cos(wt) - (1/w)(s+a) e^(-(s+a)t) cos(wt) dt

Let u = e^(-(s+a)t), dv = cos(wt) dt
and du = -(s+a) e^(-(s+a)t) dt, v = (1/w) sin(wt).

So, I = (-1/w) e^(-(s+a)t) cos(wt) - (1/w)(s+a) [(1/w) e^(-(s+a)t) sin(wt) -
(-1/w)(s+a) e^(-(s+a)t) sin(wt) dt]

Simplifying, we obtain
I = (-1/w) e^(-(s+a)t) cos(wt) - (1/w^2) (s+a) [e^(-(s+a)t) sin(wt) + (s+a) * I]
==> I = (-1/w) e^(-(s+a)t) cos(wt) - (1/w^2) (s+a) e^(-(s+a)t) sin(wt) - (1/w^2) (s+a)^2 I

Solve for I:
[1 + (1/w^2) (s+a)^2] I = (-1/w) e^(-(s+a)t) cos(wt) - (1/w^2) (s+a) e^(-(s+a)t) sin(wt)
==> [w^2 + (s+a)^2] I/w^2 = (-1/w^2) e^(-(s+a)t) [w cos(wt) + (s+a) sin(wt)]
==> I = -e^(-(s+a)t) [w cos(wt) + (s+a) sin(wt)] / [(s+a)^2 + w^2].
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Therefore, L {e^(-at) sin(wt)}
= (t = 0 to ) e^(-st) * e^(-at) sin(wt) dt
= lim(x) (t = 0 to x) e^(-st) * e^(-at) sin(wt) dt
= lim(x) -e^(-(s+a)t) [w cos(wt) + (s+a) sin(wt)] / [(s+a)^2 + w^2] {for t = 0 to x}
= lim(x) {-e^(-(s+a)x) [w cos(wx) + (s+a) sin(wx)] + w} / [(s+a)^2 + w^2]
= {0 + w} / [(s+a)^2 + w^2], by the Squeeze Theorem
= w / [(s+a)^2 + w^2].

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