Fourier Sine Transform of the derivative of a function [closed]

I'm going over some old stuff of Fourier transforms, and came across the identity $\mathscr{F}_s[f']=-\omega\mathscr{F}_c[f].$ I know this is done using integration by parts but I'm having a problem working this out.

closed as off-topic by John B, Semiclassical, MathOverview, Adam Hughes, Takumi MurayamaMay 1 '16 at 17:35

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Let $f$ be a function defined for $x\ge 0$ and $f(x)\to 0$ as $x\to\infty$. Then \begin{align} \mathscr F_s[f'(x)]&=\sqrt{\frac{2}\pi}\int_0^\infty f'(x)\sin(\omega x)\,\mathrm d x\\ &=\left.\sqrt{\frac{2}\pi}f(x)\sin(\omega x)\right|_0^\infty-\omega\sqrt{\frac{2}\pi}\int_0^\infty f(x)\cos(\omega x)\,\mathrm d x\\ &=-\omega\mathscr F_c[f(x)] \end{align}