Compute $\int_0^{2\pi} \frac{dx}{3-2\cos x + \sin x}$

Question

I am being asked to compute the integral $\int_0^{2\pi} \frac{dx}{3-2\cos x + \sin x}$.
First I wrote $$\int_0^{2\pi} \frac{dx}{3-2\cos x + \sin x}=2\int_0^{2\pi}\frac{ie^{ix}dx}{(1-2i)e^{2ix}+6ie^{ix}-(1+2i)}=2\int_\sigma \frac{dz}{(1-2i)z^2+6iz-(1+2i)}$$ with $\sigma:[0,2\pi]\to\mathbb{C}, t\mapsto e^{it}$ being the unit circle.
The singularities $\alpha_1,\alpha_2$ are found by solving $(1-2i)z^2+6iz-(1+2i)=0$, thus we find $$\alpha_1 = 2-i, \alpha_2=\frac{2}{5}-\frac{1}{5}i.$$ Now note that $\alpha_1$ is NOT in the unit circle, though $\alpha_2$ is. We find that $$\int_0^{2\pi} \frac{dx}{3-2\cos x + \sin x} = 2\bigg( 2\pi i \cdot\text{Res}_{\alpha_2}f \bigg).$$ Using the limit definition of the residue gives $\text{Res}_{\alpha_2}f=\lim_{z\to\alpha_2}(z-\alpha_2)\dfrac{1}{(z-\alpha_1)(z-\alpha_2)}=\dfrac{1}{\alpha_2-\alpha_1}=\dfrac{5}{4i-8}$.
This gives us $$\int_0^{2\pi} \frac{dx}{3-2\cos x + \sin x} = 4\pi i \cdot\text{Res}_{\alpha_2}f=\dfrac{20\pi}{4-\frac{1}{i}}$$ and I don't see how I can get rid of that $i$.
Was there something I did wrong from the start, did I make a silly mistake somewhere in the middle or am I just really close to the answer?

Answer 1

Let $R(x,y)=\dfrac1{3-2x+y}$. Then your integral is$$\int_0^{2\pi}R(\cos\theta,\sin\theta)\,\mathrm d\theta.\tag1$$Now, let\begin{align}f(z)&=\frac1zR\left(\frac{z+z^{-1}}2,\frac{z-z^{-1}}{2i}\right)\\&=\frac2{-(2+i) z^2+6 z-(2-i)}.\end{align}Then $(1)$ is equal to$$\frac1i\oint_{|z|=1}f(z)\,\mathrm dz.\tag2$$The function $f$ has two poles, at $2-i$ and at $\frac25-\frac i5$, of which only the second one is located at the open unit disk. So, by the residue theorem $(2)$ is equal to $\displaystyle2\pi\operatorname{res}_{z=2/5-i/5}f(z)=\pi$.

Answer 2

You forgot to divide by $1-2i$ since the quadratic denominator isn't monic, so the integral is$$\frac{4\pi i}{1-2i}\frac{5}{4i-8}=\frac{5\pi}{(1-2i)(1+2i)}=\pi.$$I'm not sure how you got a $4+i$ denominator.