Calculate the following integral using the residue theorem.
$$\int_{0}^{2\pi} \frac{\sin(\theta) + \cos(2\theta)}{2 + \sin(2\theta)}d\theta$$
This was my attempted method:
Let $z=e^{i\theta}$, $dz=izd\theta$
$\sin(\theta)=\frac{1}{2i}(z-\frac{1}{z})$, $\sin(2\theta)=\frac{1}{2i}(z^2-\frac{1}{z^2})$, $\cos(2\theta)=\frac{1}{2}(z^2+\frac{1}{z^2})$
By substitution,
$$\oint_{|z|=1}\frac{\frac{1}{2i}(z-\frac{1}{z})+\frac{1}{2}(z^2+\frac{1}{z^2})}{2+\frac{1}{2i}(z^2-\frac{1}{z^2})}\frac{dz}{iz}=
\oint_{|z|=1}\frac{z^4-iz^3+iz+1}{z(z^4+4iz^2-1)}dz
$$
From the denominator, we have poles at $z=0, \sqrt{\left( 2-\sqrt3 \right)}\angle-\frac{\pi}{4}, \sqrt{\left( 2-\sqrt3 \right)}\angle\frac{3\pi}{4}, \sqrt{\left( 2+\sqrt3 \right)}\angle-\frac{\pi}{4}, \sqrt{\left( 2+\sqrt3 \right)}\angle\frac{3\pi}{4}$
Only $z=0,
\sqrt{\left( 2-\sqrt3 \right)}\angle-\frac{\pi}{4},
\sqrt{\left( 2-\sqrt3 \right)}\angle\frac{3\pi}{4}$ are located within the contour path $|z|=1$.
Rewriting the denominator,
$$z(z^4+4iz^2-1)=z(z-\sqrt{\left( 2-\sqrt3 \right)}\angle-\frac{\pi}{4})(z-\sqrt{\left( 2-\sqrt3 \right)}\angle\frac{3\pi}{4})(z^2+(2+\sqrt3)i)$$
Around $z=0$,
$$\oint_{|z|=1}\frac{1}{z}\frac{z^4-iz^3+iz+1}{z^4+4iz^2-1}=2\pi i(-1)$$
Around $z=\sqrt{\left( 2-\sqrt3 \right)}\angle-\frac{\pi}{4}$,
$$\oint_{|z|=1}\frac{1}{(z-\sqrt{\left( 2-\sqrt3 \right)}\angle-\frac{\pi}{4})}\frac{z^4-iz^3+iz+1}{z(z-\sqrt{\left( 2-\sqrt3 \right)}\angle\frac{3\pi}{4})(z^2+(2+\sqrt3)i)}\approx2\pi i(0.59386\angle5.1039^\circ)$$
Around $z=\sqrt{\left( 2-\sqrt3 \right)}\angle\frac{3\pi}{4}$,
$$\oint_{|z|=1}\frac{1}{(z-\sqrt{\left( 2-\sqrt3 \right)}\angle\frac{3\pi}{4})}\frac{z^4-iz^3+iz+1}{z(z-\sqrt{\left( 2-\sqrt3 \right)}\angle-\frac{\pi}{4})(z^2+(2+\sqrt3)i)}\approx2\pi i(0.699\angle-4.3336^\circ)$$
Adding them up, the resulting sum is $\approx2\pi i(0.288675)$, which does not make sense since this is a real integral to begin with. At which step did I make a mistake?
Best Answer
Your mistake boils down to an error at some point in your arithmetic. By the Cauchy integral formula, you should have
$$\begin{align*} \oint\limits_{|z|=1} \frac{\frac{z^4-iz^3+iz+1}{z \left(z - r\,e^{i3\pi/4}\right) \left(z^2 + i\left(2+\sqrt3\right)\right)}}{z-r\,e^{-i\pi/4}} \, dz &= i2\pi \frac{r^4e^{-i\pi} - ir^3e^{-i3\pi/4} + ire^{-i\pi/4}+1}{r\,e^{-i\pi/4} \left(r\,e^{-i\pi/4} - r\,e^{i3\pi/4}\right) \left(r^2e^{-i\pi/2} + i\left(2+\sqrt3\right)\right)} \\ &= \frac{i\pi}{r^2\left(2+\sqrt3-r^2\right)} \left(1+\frac{1+i}{\sqrt2}r - \frac{1-i}{\sqrt2}r^3-r^4\right) \\[2ex] \oint\limits_{|z|=1} \frac{\frac{z^4-iz^3+iz+1}{z \left(z - r\,e^{-i\pi/4}\right) \left(z^2 + i\left(2+\sqrt3\right)\right)}}{z-r\,e^{i3\pi/4}} \, dz &= \frac{i\pi}{r^2\left(2+\sqrt3-r^2\right)}\left(1-\frac{1+i}{\sqrt2}r+\frac{1-i}{\sqrt2}r^3-r^4\right)\end{align*}$$
where $r=\sqrt{2-\sqrt3}$, and together these sum to $i2\pi$ to cancel the other residue at $z=0$.