$$\int_{0}^{2\pi} \frac{1}{a+b\sin(x)} dx = \frac{2\pi}{\sqrt{a^2-b^2}} ~ \text{if} ~ a^2 > b^2 $$
I know the trick substituting $y=\tan(x/2)$ But I'd like to know another method.
For example using COMPLEX INTEGRAL... (but I did not find yet.)
Could you give me some source? Thank you in advance.
Hint:
Use: $z=e^{ix}$, $dx=\frac{1}{i z} dz$.
You get an rational function and the contour of integration is now the unit circle. Utilizing Residue Theorem you will get the stated result.
Can you take it from here?