Integration of $e^{it}$ I am pretty sure that
$$\bigg|\int_{A}e^{it}\,dt\bigg|\leq2$$
for every measurable set $A\subseteq[-\pi,\pi]$,
but I cannot prove this...
 A: We avoid arguments that use the modulus (i.e. triangle inequality) since for some measurable sets, for instance $A = [-\pi, \pi]$, we have an overestimate $$\Big|\int_A e^{it}dt\Big| \leq 2 < \int_A|e^{it}|dt = 2\pi$$
Simply by rotating the value of the integral (which is a complex number) back to the real numbers. 
Let $f(t) = e^{it}$. Since $\int_A f dt$ is a complex number, it has magnitude and phase $$\int_A f dt = \Big|\int_A f dt\Big| e^{i\theta}$$ for some $\theta \in [-\pi, \pi]$. Note that $\theta$ is independent of time, hence
$$\Big|\int_A f dt\Big| = \int_A f e^{-i\theta} dt$$ 
The integrals are real valued, hence we consider the real component only $$\int_A f e^{-i\theta} dt = \int_A \mathfrak{Re}(f e^{-i\theta}) dt = \int_A \cos(t - \theta) dt$$
It remains to prove that the real integral $ \int_A \cos(t - \theta) dt \leq 2$.
Hint. $\cos(t-\theta)$ is non-negative for $t-\theta \in [-\pi/2, \pi/2]$, so we are interested in such elements of $A$, denote them by $A^+$, so that $$\int_A  \cos(t-\theta) dt \leq \int_{A^+} \cos(t-\theta) dt \leq \int_{[-\pi/2, \pi/2]} \cos(t)dt = 2$$
A: We have
$$ I=\int_a^be^{it}dt=\frac{1}{i}(e^{ib}-e^{ia}) $$
Then
$$ |I|=|-i||e^{ib}-e^{ia}|=|e^{ib}||1-e^{i(b-a)}|\le |1|+|e^{i(b-a)}|\le 1+1=2$$
A: Hint: $e^{it}=\cos t + i \sin t$ and split $[-\pi,\pi]$ into the regions $[-\pi,-\pi/2],[-\pi/2,0],[0,\pi/2],[\pi/2,\pi]$. How do $\cos$ and $\sin$ behave on these regions? 
