What is behavior of that integral? Does the improper integral $$\int\limits_{2\pi}^\infty \frac {\sin x} {\cos x +\ln x}\,dx $$ converge or diverge?
The Maple code $$VectorCalculus:-int(sin(x)/(cos(x)+ln(x)), x = Pi .. infinity, numeric, epsilon = .1) $$ says nothing about that.
 A: Let
$$I_m = \int_{m\pi}^{(m+1)\pi} \frac{\sin x}{\cos x + \ln x}\,dx.$$
Since $\ln x \to\infty$ for $x\to \infty$, we have $I_m\xrightarrow{m\to\infty} 0$. Furthermore, $I_m = (-1)^m\lvert I_m\rvert$.
Now it remains to check the monotonicity of $\lvert I_m\rvert$. We have
$$
\lvert I_{m-1}\rvert - \lvert I_{m}\rvert
= \int_0^\pi \frac{\sin t}{\cos (m\pi-t) + \ln (m\pi-t)}\,dt - \int_0^\pi \frac{\sin t}{\cos (m\pi+t)+\ln (m\pi+t)}\,dt
$$
and since
$$\bigl(\cos (m\pi+t)+\ln (m\pi+t)\bigr) - \bigl(\cos (m\pi-t)+\ln(m\pi-t)\bigr) = \ln (m\pi+t) - \ln (m\pi-t) > 0,$$
the monotonicity of $\lvert I_m\rvert$ follows. Now Leibniz' criterion says
$$\sum_{m=2}^\infty I_m$$
converges, and since $I_m\to 0$, we also deduce the existence of
$$\lim_{T\to\infty} \int_{2\pi}^T \frac{\sin x}{\cos x+\ln x}\,dx.$$
If $2k\pi \leqslant T \leqslant (2k+1)\pi$, then
$$\sum_{m=2}^{2k-1} I_m \leqslant \sum_{m=2}^{2k-1} I_m + \int_{2k\pi}^T \frac{\sin x}{\cos x+\ln x}\,dx = \int_{2\pi}^T\frac{\sin x}{\cos x+\ln x}\,dx \leqslant \sum_{m=2}^{2k} I_m,$$
and similarly, if $(2k+1)\pi \leqslant T \leqslant (2k+2)\pi$, then
$$\sum_{m=2}^{2k} I_m \geqslant \sum_{m=2}^{2k} I_m + \int_{(2k+1)\pi}^T \frac{\sin x}{\cos x+\ln x}\,dx = \int_{2\pi}^T \frac{\sin x}{\cos x+\ln x}\,dx \geqslant \sum_{m=2}^{2k+1} I_m.$$
Letting $S_n = \sum\limits_{m=2}^n I_m$, we have $S_2 > S_4 > \dotsc > S_{2k} > S_{2k+2} > S_{2k+3} > S_{2k+1} > \dotsc > S_3$, and hence
$$S_{2k-1} \leqslant \int_{2\pi}^T \frac{\sin x}{\cos x+\ln x}\,dx \leqslant S_{2k}$$
whenever $T \geqslant 2k\pi$. Therefore
$$\left\lvert \int_{T_1}^{T_2} \frac{\sin x}{\cos x+\ln x}\,dx\right\rvert \leqslant S_{2k} - S_{2k-1} = I_{2k}$$
whenever $T_1,T_2 \geqslant 2k\pi$.
A: The denominator is positive for all $x \in [2 \pi,\infty)$, since $\ln(x) > 1 \geq |\cos(x)|$ for $x>e$ and $2 \pi > e$. The numerator alternates in sign at each multiple of $\pi$. So the integral can be written as
$$\int_{2 \pi}^\infty \frac{\sin(x)}{\cos(x) + \ln(x)} dx = \sum_{k=2}^\infty \int_{k \pi}^{(k+1) \pi} \frac{\sin(x)}{\cos(x) + \ln(x)} dx$$
For even $k$, the numerator is positive, so we get a positive term; for odd $k$, the numerator is negative, so we get a negative term. Then you apply the alternating series test as Lucian suggested.
