Symmetry of trig integrals = $0$? $\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}$ $$\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}dx$$

Because the integral limits go from $-\pi$ to $\pi$ ...can one immediately conclude this integral equals zero? The area under the curve should cancel itself out, across the $y$-axis, right?
You can certainly use symmetry, since if
$$
f(x) = \frac{\cos(x)}{\sqrt{4+3\sin(x)}},$$
then
$$
f(x-\pi/2) = \frac{\sin (x)}{\sqrt{4-3 \cos (x)}},$$
which is an odd function. Thus,
$$
\int_{-\pi}^{\pi} f(x-\pi/2) = 0.
$$
Geometrically, this means that $f$ has odd symmetry about the line $x=\pi/2$, thus
$$
\int_{-\pi/2}^{3\pi/2} f(x) dx = 0,
$$
as you can see:



Of course, we then get that the integral is zero over any interval of length $2\pi$ by periodicity.



You can also use periodicity directly since, with substitution $u=\sin(x)$, we get
$$
\int_{-\pi}^{\pi} \frac{\cos(x)}{\sqrt{4+3\sin(x)}} \, dx =
 \int_0^0 \frac{1}{\sqrt{4+3u}} \, du = 0.
$$
In fact, this shows that 
$$
\int_I g(\sin(x))\cos(x) \, dx = 0,
$$
for any integrable function $g$ over any interval $I$ of length $2\pi$.
If $f(x)$ is an odd integrable  function and $\alpha $ is a real number then we have $$ \int _{-\alpha} ^ \alpha  f(x) dx =0$$

Unfortunately in the case of your function,  $$ f(x) = \frac {\cos x}{\sqrt {4+3\sin x}}$$ is not an odd function.

The integral may or may not be zero, but you can not use the idea of an odd function here.