Could someone please help with the following:

Find the laplace transform of the following:

$\displaystyle 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

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

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

and

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

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

skipping a few lines of calculation....

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

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

but I am not really sure. Im lost