I am calculating the following integral:
$$\int_{0}^{2\pi} \frac{1}{5-4\sin(x)} \,dx$$
So far I have the following:
Let $\sin(x) = \frac{1}{2i}$$(z – \frac{1}{z})$
So the integral becomes:
$\oint\limits_{\Gamma} \frac{1}{2z^2 + i5z-2} \,dz$
The zeroes of the denominator are $z = -2i, -\frac{i}{2}$
Since the first zero is out of the contour $\Gamma$ (the unit circle), then we exclude it. Hence, the only singularity is the second. It is a simple pole.
So, we use the residue theorem to say that the value is $2\pi i$$\operatorname{Res}f(z)$ at $z= -\frac{i}{2}$.
This is where I get stuck and get the wrong answer. So after application of the residue theorem, I get $2\pi i$$\operatorname{Res}f(z)$ = $2\pi i$$\frac{1}{z+2i}$ evaluated at $z = \frac{-i}{2}$, so final answer I get is $\frac{4\pi}{3}$. However, one can check with any software that it is half of that, or $\frac{2\pi}{3}$.
Question is, what am I missing?
Best Answer
Note that $2z^2 + 5iz - 2 = 2(z+2i)(z+\frac{i}{2})$, the $2$ factor shouldn’t be forgotten. So, we have $$2\pi i .\text{Res(f(z))} = 2\pi i \frac{1}{2(2i - \frac{i}{2})} = ??$$