Recently I posted a question regarding computation of an interesting definite integral that involved trigonometric functions. Ever since, the topic of trigonometric integrals has quite attracted me, and I decided to sift through some calculus books that I have in search of interesting and challenging problems to solve.
Yesterday I stumbled upon one that required me to compute the following integral:
$$
\mathcal{I}(0, 2\pi) = \int\limits_0^{2\pi}\frac{\sin(x)+1}{\cos(x) +2}dx
$$
It was quite challenging for me to deal with this integral, so I decided to refer to WolframAlpha, which suggested I use the substitution $u=\tan(x/2)$ to attempt the integral. And it seems like the final answer should be $\mathcal{I}(0, 2\pi)=\dfrac{2\pi\sqrt{3}}{3}$.
Surely, since the function under the integral is defined everywhere in $\mathbb{R}$ (with period $\mathrm{lcm}(2\pi,2\pi)=2\pi$) the above substitution proposal seems valid. Yet I wonder whether there is some other “smarter” analytical approach to deal with this integral apart from tangential substitution.
Any ideas or suggestions will be greatly appreciated.
Best Answer
Decompose the integrand as follows \begin{align} & \int_0^{2\pi}\frac{\sin x+1}{\cos x +2}\ dx\\ =& \int_ 0^{2\pi}\left( \frac{\sin x}{\cos x +2}- \frac{\cos x+2-\sqrt3}{\sqrt3(\cos x +2)}+\frac1{\sqrt3}\right)dx\\ =& \bigg[\ln(\cos x+2)-2\tan^{-1}\frac{\sin x}{\cos x+2+\sqrt3}+\frac x{\sqrt3}\bigg]_0^{2\pi}=\frac{2\pi}{\sqrt3} \end{align} where the first two terms vanish due to periodicity.