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 :)

Hint: split the integral into two pieces $-\pi$ to $0$ and $0$ to $\pi$. Note the point symmetry of the integrand for the points $P_1=(-\pi/2,0)$ and $P_2=(\pi/2,0)$.  You can achieve this by substitution $x=u\pm \pi/2$ and $dx = du$. You should see that you will get odd functions after the substitutions, which both evaluate to $0$.

An alternative approach: Use the substitution $\sin(x)=\sqrt{5}u \implies \cos(x)dx=\sqrt{5}du$:
$$\int_{-\pi}^\pi \frac{\cos(x)}{5+\sin(x)^2}dx=\frac{1}{5}\int\frac{du}{1+u^2}=\frac{1}{5}\arctan(u)=\frac{1}{5}\arctan \left(\frac{\sin(x)}{\sqrt{5}}\right)\biggr|_{-\pi}^{\pi}=0$$
