Integration of $\cos(x)/(5+\sin(x)^2)$ Is the following integration correct?
Consider the integral $\int_{-\pi}^\pi \frac{\cos(x)}{5+\sin(x)^2} dx$.
Substitute $y = \sin(x)$ then we have $\frac{dy}{dx} = \cos(x)$ and hence $dy = \cos(x) dx$ and we get
$$\int_{-\pi}^\pi \frac{\cos(x)}{5+\sin(x)^2}dx = \int_{\sin(-\pi)}^{\sin(\pi)} \frac{1}{5+y^2}dy = \int_{0}^{0} \frac{1}{5+y^2}dy = 0.$$
The substitution seems a bit odd, but the result $0$ is correct. Thanks in advance :)

You are correct that the integral is 0.
To proceed with the integration:
do the substitution 
$\sin x = \sqrt 5 \tan t\\ \cos x \;dx = \sec^2 t\; dt$ 
$\int_{-\pi}^{\pi} \frac {\cos x}{5 + \sin^2 x} dx = \frac 1{\sqrt5} \tan^{-1}(\sqrt 5 \sin x)|_{-\pi}^{\pi}$
Does that really help to explain why it integrates to 0?
perhaps not.
The integrand is periodic with period $2\pi$, i.e. $f(x+2\pi) = f(x)$ and we are integrating over the full period.
Furthermore, over a full period, the wave spends equal time below the line as it does above the line.  $f(x) = -f(x+\pi) = -f(x-\pi)$
For every point in the interval, there is a corresponding point where $f(a) = -f(b).$  And, that means that over any interval of $2\pi$ the integral will be $0.$
