# Math Help - Laplace Transform

1. ## Laplace Transform

Find the laplace transform of the following:

$f(t) = e^{4t} \ \sin^2(3t)$

I dont know if my method is correct, but here is what I started with.

f(t) becomes

$f(t) = \frac{1}{2} e^{4t} - \frac{1}{2} e^{4t} \cos(6t)$

Let $g(t) = e^{4t}, ... h(t) = e^{4t} \cos(6t)$

and
$A = \frac{1}{2}, B = -\frac{1}{2}$

Then $F(s) = \mathcal{L}(f(t)) = \int_0^{\infty} e^{-st} (A(g(t)) + B(h(t) )$

skipping a few lines of calculation....

$= \frac{1}{2} \int_0^{\infty} e^{(4-s)t} dt + (\frac{-1}{2} ) \ \mathcal{L} (\cos(6t) e^{4t})$

$= \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}$

but I am not really sure. Im lost

2. Originally Posted by Maccaman
Find the Laplace transform of the following:

$f(t) = e^{4t} \ \sin^2(3t)$

.....

$= \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}$

but I am not really sure. Im lost

3. Originally Posted by Maccaman

Find the laplace transform of the following:

$f(t) = e^{4t} \ \sin^2(3t)$

I dont know if my method is correct, but here is what I started with.

f(t) becomes

$f(t) = \frac{1}{2} e^{4t} - \frac{1}{2} e^{4t} \cos(6t)$

Let $g(t) = e^{4t}, ... h(t) = e^{4t} \cos(6t)$

and
$A = \frac{1}{2}, B = -\frac{1}{2}$

Then $F(s) = \mathcal{L}(f(t)) = \int_0^{\infty} e^{-st} (A(g(t)) + B(h(t) )$

skipping a few lines of calculation....

$= \frac{1}{2} \int_0^{\infty} e^{(4-s)t} dt + (\frac{-1}{2} ) \ \mathcal{L} (\cos(6t) e^{4t})$

$= \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}$

but I am not really sure. Im lost
You can just use Frequency shifting $\mathcal{L}\left\{ {{e^{\alpha t}}f\left( t \right)} \right\} = F\left( {s - \alpha } \right)$

So $g\left( t \right) = {e^{4t}}{\sin ^2}3t$

Then $f\left( t \right) = {\sin ^2}3t$

$\mathcal{L}\left\{ {{{\sin }^2}3t} \right\} = \mathcal{L}\left\{ {\frac{1}{2}\left( {1 - \cos 6t} \right)} \right\} = \frac{1}{2}\left( {\frac{1}{s} - \frac{s}{{{s^2} + 36}}} \right)$

Finally $\mathcal{L}\left\{ {{e^{4t}}{{\sin }^2}3t} \right\} = \frac{1}{2}\left( {\frac{1}{{s - 4}} - \frac{{s - 4}}{{{{\left( {s - 4} \right)}^2} + 36}}} \right)$