Complex integration exercise

Question

I have to calculate $$\int_0^{2\pi}\arctan \biggl( \frac{\sin \theta} {\cos \theta +3}\biggr)d\theta$$ I don't have any ideas, but I think that I should apply a substitution. Can you give me a hint? Thank you in advance

Answer 1

$$\arctan\left(\frac{\sin\theta}{\cos\theta + 3}\right)=\text{Im}\log\left(i\sin\theta+\cos\theta+3\right)=\text{Im}\log(3+e^{i\theta})=\text{Im}\log\left(1+\frac{e^{i\theta}}{3}\right) $$ so the integral equals $$ \text{Im}\int_{0}^{2\pi}\sum_{n\geq 1}(-1)^{n+1}\frac{e^{ni\theta}}{n 3^n}\,d\theta=\text{Im}(0)=\color{red}{0}.$$

Answer 2

oops, I missed the part it should be done by complex integration... Anyway, I'll leave this result as a hint what to look for. Note that $$ \arctan\left(\frac{\sin(\pi-x)}{3+\cos(\pi-x)}\right) = \arctan\left(\frac{\sin(x)}{3-\cos(x)}\right) $$ and $$ \arctan\left(\frac{\sin(\pi+x)}{3+\cos(\pi+x)}\right) = -\arctan\left(\frac{\sin(x)}{3-\cos(x)}\right) $$ thus integration from $0$ to $2\pi$ yields $0$.