Convergence of an improper integral $\int_3^\infty \frac{\sin(x)}{x+2\cos(x)}dx$ I tried to check whether the following integral converges:
$$\int_3^\infty \dfrac{\sin(x)}{x+2\cos(x)}dx$$
Dirichlet criterion doesn't work here since the function 
$$\dfrac{1}{x+2\cos(x)}$$
is not monotone.

\begin{align*}
\int_{3}^{M}\dfrac{\sin x}{x+2\cos x}dx&=-\dfrac{\cos x}{x+2\cos x}\bigg|_{x=3}^{x=M}-\int_{3}^{M}\dfrac{-(2\sin(x)-1)\cos x}{-(x+2\cos x)^{2}}dx\\
&=-\dfrac{\cos M}{M+2\cos M}+\dfrac{\cos 3}{3+2\cos 3}-\int_{3}^{M}\dfrac{-\sin(2x)+\cos(x)}{(x+2\cos x)^{2}}dx.
\end{align*}
Now each of the integrals
\begin{align*}
\int_{3}^{\infty}\dfrac{-\sin(2x)}{(x+2\cos x)^{2}}dx.
\end{align*}
\begin{align*}
\int_{3}^{\infty}\dfrac{\cos(x)}{(x+2\cos x)^{2}}dx.
\end{align*}
absolutely converges, since 
\begin{align*}
\int_{3}^{M}\dfrac{1}{(x+2\cos x)^{2}}dx\leq\int_{3}^{M}\dfrac{1}{(x-2)^{2}}dx.
\end{align*}
So the improper integral converges.
