$|e^{2x\pi i}-1|\leq 2\pi |x|$

Show $|e^{2x\pi i}-1|\leq 2\pi |x|$ for real $x$.

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We first prove another result: $$\bigg| \int_b^a f(t) dt\bigg|\leq \int_b^a |f(t)| dt.$$ $$\bigg| \int_b^a f(t) dt\bigg|\stackrel{(*)}{=}e^{-i\phi}\int_b^a f(t) dt=\int_b^a e^{-i\phi}f(t) dt$$ $(*)\quad F=|F|e^{i\phi}$ and $e^{i\phi}e^{-i\phi}=1$ means $|F|=Fe^{-i\phi}$.
Since $\bigg| \int_b^a f(t) dt\bigg|$ is real and $\text{Re}(z)\leq |z|$, we have $$\begin{align}\bigg| \int_b^a f(t) dt\bigg|&=\text{Re}\bigg(\int_b^a e^{-i\phi}f(t) dt\bigg)\\&=\int_b^a \text{Re}(e^{-i\phi}f(t)) dt\\&\leq \int_b^a |e^{-i\phi}f(t)| dt\\&=\int_b^a |f(t)| dt.\end{align}$$ One can also prove this using Cauchy Schwarz inequality.

Onto the problem, let $f(t)=e^{ixt}$ for $x,t\in \Bbb R$. $$\bigg|\int_0^{2\pi}e^{ixt}dt\bigg|\leq \int_0^{2\pi}|e^{ixt}|dt=2\pi$$ Also $$\bigg|\int_0^{2\pi}e^{ixt}dt\bigg|=\bigg[\bigg|\frac{e^{ixt}}{ix}\bigg|\bigg]_0^{2\pi}=\frac{|e^{2x\pi i}-1|}{|x|}.$$ Therefore $$|e^{2x\pi i}-1|\leq 2\pi |x|.$$