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?
 A: 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 $2\pi$.
A: The integrand is not odd and in fact is not symmetric, but
$$\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}=\frac23\sqrt{4+3\sin(x)}\Big|_{-\pi}^{\pi}=0$$
A: Using basic trigonometric identities and the substitution $x\to-x$,
$\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx\\= \int_{-\pi}^{-\pi/2}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx+\int_{-\pi/2}^{0}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx+\int_{0}^{\pi/2}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx+\int_{\pi/2}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx\\= \int_{0}^{\pi/2}\frac{\cos(x-\pi)}{\sqrt{4+3\sin(x-\pi)}}\,dx+\int_{\pi/2}^{0}\frac{-\cos(-x)}{\sqrt{4+3\sin(-x)}}\,dx+\int_{0}^{\pi/2}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx+\int_{-\pi/2}^{0}\frac{\cos(x+\pi)}{\sqrt{4+3\sin(x+\pi)}}\,dx
\\= \int_{0}^{\pi/2}\frac{-\cos(x)}{\sqrt{4-3\sin(x)}}\,dx+\int_{\pi/2}^{0}\frac{-\cos(x)}{\sqrt{4-3\sin(x)}}\,dx+\int_{0}^{\pi/2}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx+\int_{-\pi/2}^{0}\frac{-\cos(x)}{\sqrt{4-3\sin(x)}}\,dx\\
= \left(\int_{0}^{\pi/2}\frac{-\cos(x)}{\sqrt{4-3\sin(x)}}\,dx+\int_{\pi/2}^{0}\frac{-\cos(x)}{\sqrt{4-3\sin(x)}}\,dx\right)+\left(\int_{0}^{\pi/2}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx+\int_{\pi/2}^{0}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}\,dx\right)\\=0+0=0.$
A: 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.    
