How to identify quick ways of evaluating integrals of complex exponentials? For example, on the interval $-\pi\le x\le \pi$ consider $$\int_{x=-\pi}^{\pi}\Big(\left(e^{3ix}-e^{-3ix}\right)-\left(e^{ix}-e^{-ix}\right)\Big)dx$$ 
One answer states that:

$$\int_{x=-\pi}^{\pi}\Big(\left(e^{3ix}-e^{-3ix}\right)-\left(e^{ix}-e^{-ix}\right)\Big)dx=0\tag{1}$$ The equality follows as the complex exponentials are integrated over $3$ and $1$ full period, respectively, and thus average out to zero.

I'm confused by what the author means by 'integrated over $3$ full periods'. It was my understanding that one full period is $2\pi$. So $3$ full periods must be $6\pi$ so the limits should be $-3\pi\le x\le 3\pi$. 
Could someone please explain how I can effectively 'see' that this integral is zero, without going through the calculation? 
Using Euler's formula I write $$e^{3ix}=\cos(3x)+i\sin(3x)$$
in order to better understand what the author is talking about, but I'm still confused. 
Does anyone have any idea what the author means by this part:

and thus average out to zero


After an online search I found that another answer gave the following reasoning from this website:

For an odd function $$f(x)=-f(-x):\int_a^af(x)dx=0$$

Now I have no idea whatsoever what this is supposed to mean.
If someone could shed some light on why equation $(1)$ is zero it would be greatly appreciated. 
Thank you.
 A: Your internet research brought you on the right track. Yet the formula is somewhat different.

For an odd, integrable function $f:[-a,a]\rightarrow\mathbb{R}$ the following is valid
  \begin{align*}
\int_{-a}^a f(x)\,dx = 0
\end{align*}

Recall a function $f$ is odd iff
\begin{align*}
f(x)=-f(-x)\tag{1}
\end{align*}
On the other hand we know the integration rules
\begin{align*}
\int_a^b f(x)\,dx&=-\int_b^a f(x)\,dx\tag{2}\\
\int_{a}^b f(x)\,dx&=\int_{a}^c f(x)\,dx+\int_{c}^bf(x)\,dx\tag{3}\\
\int_{a}^b f(x)\,dx&=-\int_{-a}^{-b}f(-x)\,dx\tag{4}
\end{align*}
The last rule (4) is due to substitution
\begin{align*}
x&\rightarrow -x\\
dx&\rightarrow -dx
\end{align*}

Therefore we obtain for odd functions $f:[-a,a]\rightarrow\mathbb{R}$
\begin{align*}
\int_{-a}^a f(x)\,dx&=\int_{-a}^0 f(x)\,dx + \int_{0}^a f(x)\,dx\qquad\qquad&\longrightarrow (3)\\
&=-\int_{0}^{-a}f(x)\,dx + \int_{0}^a f(x)\,dx\qquad\qquad&\longrightarrow (2)\\
&=\int_{0}^{a}f(-x)\,dx + \int_{0}^a f(x)\,dx\qquad\qquad&\longrightarrow (4)\\
&=-\int_{0}^{a}f(x)\,dx + \int_{0}^a f(x)\,dx\qquad\qquad&\longrightarrow (1)\\
&=0
\end{align*}
Since $e^{3ix}-e^{-3ix}$ is odd as well as $e^{ix}-e^{-ix}$ we can see at a glance
  \begin{align*}
\int_{-\pi}^\pi\left(\left(e^{3ix}-e^{-3ix}\right)-\left(e^{ix}-e^{-ix}\right)\right)=0
\end{align*}

$$ $$

Regarding integration over $3$ full periods: Let's consider
  \begin{align*}
&e^{ix}\qquad\qquad &-6\pi\leq x\leq 6\pi\\
&e^{3ix}\qquad\qquad &-2\pi \leq x\leq 2\pi
\end{align*}
  
  
*
  
*In the first case we have three full periods by increasing the length of the interval by a factor three from $2\pi$ to $6\pi$.
  
*In the second case we have three full periods by increasing the speed of rotation by a factor three from $e^{ix}$ to $e^{3ix}$.
In both cases we have three full periods.

A: You were on the right track using Euler's formula. Note that $\cos{3x}$ and $\sin{3x}$ both have period $\frac{2\pi}{3}$, so 3 full periods is indeed $2\pi$. 
As for the second part, notice $f(x) = e^{aix} - e^{-aix}$ is odd since $-f(-x) = -(e^{ai(-x)} - e^{-ai(-x)}) = -(e^{-aix} - e^{aix}) = f(x)$. In fact, using Euler's formula as you gave shows $f(x) = 2i \sin{ax}$, from which you can deduce that $f$ is odd directly. 
So we have the sum of two odd functions, which is also an odd function. Does this all make sense?
EDIT: 
Your integral now comes to this. 
$$ \int_{-\pi}^{\pi} 2i \sin{3x} - 2i \sin{x} \: dx
 = \int_{-\pi}^{\pi} 2i \sin{3x} \: dx - \int_{-\pi}^{\pi} 2i \sin{x} \: dx$$
Both of those functions are odd, so each individual integral is $0$.
Here's the other way to think about it. Consider the integral we just looked at, but pull out the constant $2i$ (just for clarity). 
$$ 2i \int_{-\pi}^{\pi} \sin{3x} \: dx - 2i \int_{-\pi}^{\pi} \sin{x} \: dx $$
Think of the interpretation of the integral as the (signed) area under the curve. The overall area under the curve, for one full period of the $\sin$ function is $0$. Since the two intgrands are making 3 and 1 full periods respectively, the values of the integrals are  $3*0 = 0$ and $0$ respectively. 
