Compute the integral $$ \int_0^{2\pi} \frac{1 + \sin(\theta)}{3 + \cos(\theta)}d \theta$$
My attempt: Using residue integration and rewriting the expression using complex identities for $\sin(\theta)$ and $\cos(\theta)$: $$\sin(\theta) = \frac{1}{2i}\left(z – \frac{1}{z}\right), \hspace{4mm} \cos(\theta) = \frac{1}{2}\left(z + \frac{1}{z}\right)$$ and $d\theta = \frac{dz}{iz}$:
$$\int_0^{2\pi} \frac{1 + \sin(\theta)}{3 + \cos(\theta)}d \theta$$ $$= \int_C \frac{1 + \frac{1}{2i}(z – \frac{1}{z})}{3 + \frac{1}{2}(z+ \frac{1}{z})} \frac{dz}{iz}$$ $$= \frac{1}{2i} \int_C \frac{1 + \frac{1}{2i}(z – \frac{1}{z})}{(z + 3 – 2\sqrt{2})(z + 3 + 2\sqrt{2}) }dz$$ We see that only the singularity $z = 2 \sqrt{2} – 3$ lies inside the unit circle, and evaluating this singularity gives
$$r = \frac{1 + \frac{1}{2i}(z – \frac{1}{z})}{z + 3 + 2\sqrt{2}}_{z = 2\sqrt{2} – 3} = \frac{1}{4\sqrt{2}} – i/2$$ which gives the final answer:
$2\pi i(\frac{1}{2i})(\frac{1}{4\sqrt{2}} – i/2) = \frac{\pi}{4 \sqrt{2}} – \frac{\pi i}{2}$ whereas the answer should be $\pi/\sqrt{2}$
Best Answer
The integrand that you have written, $$ \frac{1+\frac 1{2 i}(z-\tfrac{1}{z})}{3+\frac 12(z+\tfrac{1}{z})}\frac {\mathrm d z}{i z} = -\frac{(z+i)^2}{z (z+3-2\sqrt 2)(z+3+2\sqrt 2)}\mathrm dz, $$ is correct; however its poles within $|z|<1$ are $z=2\sqrt 2-3$ and $z=0$. You are on the right track with the first one, but you also need to add the residue from $z=0$.