Residue Integration

Question

I am attempting to calculate the integral of $\frac{(1+sin(\theta))}{(3+cos(\theta))}$ from $0$ to $2\pi$. I have already changed $sin$ and $cos$ into $\frac{1}{2i(z-z^{-1})}$ and $\frac{1}{2(z+z^{-1})}$. I am really stuck now. Can anyone please guide me?

Answer 1

If direct way can be considered, just as Dr.MW already answered, tangent half-angle substitution $t=\tan \frac \theta 2$ makes the problem simple since $$I=\int\frac{(1+sin(\theta))}{(3+cos(\theta))}d\theta=\int\frac{(t+1)^2}{t^4+3 t^2+2}dt=\int \frac{2 t}{t^2+1} dt+\int\frac{1-2 t}{t^2+2}dt$$ $$I=\log \left(1+t^2\right)+\frac{\tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)}{\sqrt{2}}-\log \left(2+t^2\right)=\frac 1{\sqrt{2}}\tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+\log\Big(\frac{1+t^2}{2+t^2} \Big)$$ Back to $\theta$ (if required), $$I=\frac{1}{\sqrt 2}\tan ^{-1}\left(\frac{\tan \left(\frac{\theta }{2}\right)}{\sqrt{2}}\right)-\log (3+\cos (\theta ))$$ So, since, as explained by Dr.MW,$$ \int_0^{2\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta=\int_{-\pi}^{\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta $$ the bounds for $t$ are $-\infty$ and $+\infty$, so the logarithmic terms does not contribute and the result is just $\frac{\pi}{\sqrt 2}$.

More generally, assuming $a \leq \pi$,$$\int_{-a}^{a} \frac{1+\sin \theta}{3 + \cos \theta} d\theta=\sqrt{2} \tan ^{-1}\left(\frac{a}{\sqrt{2}}\right)$$

Answer 2

Since the integrand is periodic, then

$$\begin{align} \int_0^{2\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta&=\int_{-\pi}^{\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta\\ &=\int_{-\pi}^{\pi} \frac{1}{3 + \cos \theta} d\theta \end{align}$$

where we exploited the fact that $\frac{\sin \theta}{3+\cos \theta}$ is an odd function.

Next, let $u=\tan (\theta /2)$ so that $du = \frac12 \sec^2(\theta /2)$, and $\cos \theta =\frac{1+u^2}{1-u^2}$. Then, we find the anti-derivative of $\frac{1}{3+\cos \theta}$ is $\frac{\sqrt{2}}{2} \arctan (\sqrt{2}\tan (\theta /2)/2) +C$. Evaluating the anti-derivative between limits of integration reveals

$$\int_{-\pi}^{\pi} \frac{1}{3+\cos \theta} d\theta =\frac{\sqrt{2}\pi}{2}$$

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Now, let's use contour integration.

Let $z=e^{i \theta}$ so that $d\theta=dz/(iz)$.

Next note that

$\frac{1+\sin \theta}{3+\cos \theta}=\frac{z^2+2iz-1}{i(z^2+6z+1)}$. The only root of the denominator that lies inside the unit circle is at $z=-3+2\sqrt{2}$.

Now the integral of interest is

$$-\int_C \frac{z^2+2iz-1}{z(z^2+6z+1)} dz$$

The integral has two simple poles, one at $0$ and the other at $-3+2\sqrt{2}$. The evaluation of the integral therefore is, by the Residue Theorem,

$$2\pi i \sum \text{Res} \left(- \left( \frac{z^2+2iz-1}{z(z^2+6z+1)}\right)\right)$$

The residue at $z=0$ is found by evaluating the term $- \frac{z^2+2iz-1}{z^2+6z+1}$ at $z=0$. We find this first residue to be 1.

The residue at $z=-3+2\sqrt{2}$ is found by evaluating the term $-\frac{z^2+2iz-1}{z(z+3+2\sqrt{2})}$ at $z=-3+2\sqrt{2}$. We find this second residue to be $-1-i\sqrt{2}/4$.

Putting it all together recovers the aforementioned result!