Compute a laplace transform 
Compute $\mathcal{L} [e^{-2x} \sin x]$

So I want to do $\int_0^\infty e^{-2x} \sin x e^{-px} \, dx=\int_0^\infty e^{-(p+2)x}\sin x \, dx$
But due to the way $e$ and $\sin x$ integrates I dont know where to go from here?
 A: Try integration by parts, twice.  When you have done that you find you have gone in a circle...you should have something along the lines of:
$$f(p) - g(p)\int_0^\infty e^{-(p+2)x} \sin x \,dx$$
And you have the same integral that you already know that you didn't know how to solve... but you are not dead yet.
$$\int_0^\infty e^{-(p+2)x} \sin x \, dx = f(p) - g(p)\int_0^\infty e^{-(p+2)x} \sin x \, dx\\
(1+g(p)) \int_0^\infty e^{-(p+2)x} \sin x \, dx = f(p)\\
\int_0^\infty e^{-2x} \sin x e^{-px} \, dx = \frac {f(p)}{1+g(p)}$$
Alternatively say: $\sin x = \frac 1{2i} (e^{ix} - e^{-ix})$
$$\frac 12 \int_0^\infty e^{-(p+2-i)x} - e^{-(p+2+i)x} \, dx$$
A: Hint. One may write
$$
\int_{0}^{\infty}e^{-(p+2)x}\sin x dx=\text{Im}\int_{0}^{\infty}e^{-(p+2-i)x}\:dx=\text{Im}\:\frac1{p+2-i}=\frac1{(p+2)^2+1}, \quad p>-2.
$$
A: We have $$J=\int_0^\infty e^{-2x} \sin xe^{-px} \, dx =\int_0^\infty e^{-(p+2)x} \sin x\,dx=-\int_0^\infty e^{-(p+2)x} \, d(\cos x) =\\ =-\left( \left. e^{-(p+2)x \cos x} \right|_0^\infty +(p+2) \int_0^\infty e^{-(p+2)x} \cos x \, dx \right) =\\ =1-(p+2) \int_0^\infty e^{-(p+2)x} \cos x \, dx =1-(p+2) \left( \left. e^{ -(p+2)x \sin x } \right|_0^\infty +(p+2) \underbrace { \int_0^\infty e^{ -(p+2)x} \sin x \, dx}_J \right) $$
so the answer is

$$ J=1-(p+2)^2 J\Rightarrow \quad J=\frac 1 {1+(p+2)^2 } $$

A: I realize that you want to understand how to find the transform using the integral definition and others have already indicated ways to do that.
But for the sake of variety here is how to prove it using only linearity and the property $\mathcal{L}\{y^\prime\}=s\mathcal{L}\{y\}-y(0)$.
Let $y=e^{-2x}\sin(x)$. Then 
\begin{equation}
y^\prime=-2e^{-2x}\sin(x)+e^{-2x}\cos(x)
\end{equation}
and
\begin{equation}
y^{\prime\prime}=3e^{-2x}\sin(x)-4e^{-2x}\cos(x)
\end{equation}
So
\begin{equation}
y^{\prime\prime}+4y^\prime=-5e^{-2x}\sin(x) \tag{1}
\end{equation}
Note that $y(0)=0$ and $y^\prime(0)=1$. Take the Laplace transform of both sides.
\begin{eqnarray}
s\mathcal{L}\{y^\prime\}-y^\prime(0)+4[s\mathcal{L}\{y\}-y(0)] &=-5\mathcal{L}\{y\}\\
s[s\mathcal{L}\{y\}-y(0)]-1+4s\mathcal{L}\{y\}&=-5\mathcal{L}\{y\}\\
\left(s^2+4s+5\right)Y(s)&=1\\
Y(s)&=\dfrac{1}{(s+2)^2+1}
\end{eqnarray}
A: Hint:$$F(s-p)=\int_0^\infty{e^{-(s-p)t}f(t)}=$$
$$\int_0^\infty{e^{-st}e^{pt}f(t)}=\mathcal L(e^{pt}f(t))$$
Using this to compute $\mathcal{L} [e^{-2x} \sin x]$ gives:
$$\mathcal{L} (e^{-2x} \sin x)=\dfrac{1}{(s+2)^2+1}$$
Computing $\mathcal{L} [e^{-(p+2)x} \sin x]$ gives:
$$\mathcal{L} (e^{-(p+2)x} \sin x)=\dfrac{1}{(s+p+2)^2+1}$$
