Proving $\int_{0}^{\infty}\frac{w^3\sin(wx)}{w^4+4} \ dw=\frac{\pi}{2}e^{-x}\cos(x)$ 
I am trying to prove $$\int_{0}^{\infty}\frac{w^3\sin(wx)}{w^4+4} \ dw=\frac{\pi}{2}e^{-x}\cos(x),  \ \ x>0$$ using an appropriate transform.

I thought of using a sine transform. The inverse of the sine transform is defined as $$\mathcal{F}^{-1}_s(F(w))=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\sin(wx)F(w) \ dw.$$ Here, $$F(w)=\frac{w^3}{w^4+4}.$$
I thought I could prove the result by taking $$\mathcal{F}_s\left(\frac{\pi}{2}e^{-x}\cos(x)\right)=\sqrt{\frac{\pi}{2}}\int_{0}^{\infty}\sin(wx)e^{-x}\cos(x) \ dx,$$ but I do not know how to solve this integral. A hint would be great.
 A: I guess this might help. The Fourier Sine Integral is given by
$$
f(x) = \int_{0}^{\infty }B(w)\sin(wx)dw
$$
where
$$
B(w) = \frac{2}{\pi}\int_{0}^{\infty }f(x)\sin(wx)dx
$$
Now it is given that
$$
f(x) = \frac{\pi}{2}e^{-x}\cos(x)
$$
Therefore,
$$
B(w) = \frac{2}{\pi}\int_{0}^{\infty }\frac{\pi}{2}e^{-x}\cos(x)\sin(wx)dx
$$
$$
B(w)  = \int_{0}^{\infty }e^{-x}\cos(x)\sin(wx)dx \hspace{1cm} ... (1)
$$
Also,
$$
\sin(wx) \cos(x) = \frac{\sin((w+1)x) - \sin((w-1)x)}{2} \hspace{1cm} ... (2)
$$
Now substitute (2) in (1) to get
$$
B(w) = \frac{1}{2}\left [ \int_{0}^{\infty } e^{-x} \sin((w+1)x) dx - \int_{0}^{\infty } e^{-x} \sin((w-1)x) dx \right ]
$$
In order to solve the above integral, we need the following equation (more on this can be found here):
$$
\int{e^{ax}\sin({bx})dx}= \dfrac{e^{ax}}{a^{2}+b^{2}}[a\sin(bx)-b\cos(bx)] + C
$$
If we substitute $a = -1$ and $b = w+1$, we get:
$$
\int_{0}^{\infty } e^{-x} \sin((w+1)x) dx = \frac{w+1}{1+(w+1)^2}
$$
and when we substitute $a = -1$ and $b = w-1$, we get:
$$
\int_{0}^{\infty } e^{-x} \sin((w-1)x) dx = -\frac{w-1}{1+(w-1)^2}
$$
Therefore,
$$
B(w) = \frac{1}{2}\left [ \frac{w+1}{1+(w+1)^2} + \frac{w-1}{1+(w-1)^2} \right ] = \frac{w^3}{w^4+4}
$$
Thus,
$$
\int_{0}^{\infty}\frac{w^3\sin(wx)}{w^4+4} \ dw=\frac{\pi}{2}e^{-x}\cos(x)
$$
A: Outline.
Expanding on @Mattos' hint, an elementary technique is to exploit the identity $2\sin A\cos B=\sin(A+B)-\sin(A-B)$ so that the indefinite integral is proportional to $I_1-I_2$ where both are proportional to $\int\frac{\sin Kx}{\exp x}\,dx$ for a constant $K$. Now use integration by parts, the first time differentiating $\sin Kx$ and integrating $\exp x$ to get an integral proportional to $\int\frac{\cos Kx}{\exp x}\,dx$. Use integration by parts a second time, differentiating $\cos Kx$ and integrating $\exp x$ to get an integral proportional to $\int\frac{\sin Kx}{\exp x}\,dx$. Notice that this is the same integral as the first one, so you can collect the integral terms to solve for $\int\frac{\sin Kx}{\exp x}\,dx$.
