Is there a way to figure out which trigonometric integrals will be automatically zero? Now we have some examples of what I mean $$\int_0^{2\pi} \sin x~dx=0$$
$$\int_0^{8\pi} \cos 4x~dx=0$$
$$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$
Looking at the graph of $f(x)=\sin (x)$ for example it makes some sense to me that $$\int_0^{2\pi} \sin x~dx=0$$ because the region below the $x$ axis will "cancel" with the part above but I don't understand how I can make claims about more general integrals like  $$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$ which I just got from testing with an online integral tool.
Obviously there is something more deeper going on and I would like to understand when and why I can just claim that an integral will be zero as it will save me a lot of time.
I have had a thought but can't seem to conclude how it would be obvious that any of the less obvious integrals would be zero (like the third one I stated), I imagine that there is some kind of symmetry but I'm just not seeing it. 
So if anyone could explain when (in a more general sense not just for the integrals above) and why this works I would appreciate it. 
 A: Roughly speaking, a trig integral (sine or cosine) that includes an integer number of cycles will be zero. The same goes for $\sin nx$ or $\cos nx$. Your "area above the line is the same as area below" is a perfectly good explanation. 
As for powers of sines, any odd power of sine can be simplified by the rule 
$$
\cos 2x = 1  - 2 \sin^2 x
$$
by rewriting 
$$
\sin^2 x = \frac{1 - \cos 2x}{2}
$$
For instance, 
$$
\sin^3 x = \sin x \sin^2 x = \sin x \frac{1 - \cos 2x}{2}
$$
Using this, you can express any odd power of sines as a product of lower power sines and cosines (albeit with $2x$ or $3x$ or $4x$, etc., instead of $x$) and such things also integrate to zero be the periodicity-and-symmetry argument.
On the other hand, if you integrate $\sin (x/2)$ from $0$ to $4 \pi$, for instance, you'll get zero as well, but not if you integrate only to $2\pi$. 
That's because $2\pi$ is only a half-period for this function. 
There are other ways to get zero: integrate $\sin$ from $-a$ to $a$, for instance. These depend on the symmetry of the function rather than its periodicity, and seem somehow different form the examples you gave. 
There are other examples that are helpful to know even though you don't get zero. For instance, look at
$$
Q = \int_0^{2\pi} \sin^2(x) dx.
$$
That's clearly the same as 
$$\int_0^{2\pi} \cos^2(x) dx,$$
because $\sin$ and $\cos$ are just shifted version of each other, and we're integrating over a whole period of either one. 
But that means that 
$$
2Q = 
\int_0^{2\pi} \sin^2x + \cos^2 x ~dx= \int_0^{2\pi} 1 ~dx= 2\pi,
$$
so 
$Q = \pi$. 
And in general, integrals of even powers of sines and cosines will work out like this: they're always positive, so their integrals (over any interval of length > 0) will be positive. 
A: Hint : for the first one, use the change of variable $y = x-\pi$ on $[\pi,2\pi]$, the fact that the function $sin$ is an odd function and $$\int_0^{2\pi} \sin x~dx=\int_0^{\pi} \sin x~dx+\int_{\pi}^{2\pi} \sin x~dx$$
A: Hint
If $f$ is periodic and odd, you will have $$\int_a^{a+T}f(x)dx=0$$
for all $a\in\mathbb R$.
For $$\int_\pi^{2\pi}\sin^3(10 x)dx$$
you have that $x\longmapsto \sin^3(10x)$ is odd and $\frac{\pi}{5}-$periodic. Moreover, $$\int_\pi^{2\pi}\sin^3(10x)dx=\sum_{k=0}^4\underbrace{\int_{\pi+\frac{k\pi}{5}}^{\pi+\frac{(k+1)\pi}{5}}\sin^3(10 x)dx}_{=0\ for\ all\ k}=0.$$
A: If $p$ is a period for $f$, then $p$ is also a period for $f^n$.
If the integral is over an interval whose length is a whole number of periods for the integrand, and if the integral over a single period is zero, then that integral is also zero.
A period for $\sin 10x$ is $p=2\pi/10 = \pi/5$. So $\pi/5$ is also a period for $\sin^3 10x$. The interval of integration is of length $\pi$, which is $5\times\pi/5$, so is a whole number of periods of the integrand.
It remains to see that the integral of $f(x)\equiv\sin^3 10x$ over a single period is zero. But that follows quickly from the observation that $$f( x+\frac12 p) = (\sin 10(x+\pi/10))^3$$
$$=(\sin (10x+\pi))^3 = (-\sin 10x)^3$$
$$=-\sin^310x=-f(x)$$
since the integral over the first half of the period would then add out with the integral over the second half of the period.
