Convergence of an improper integral involving trigonometric functions 
Determine the convergence of the following improper integral:
$$\int_{2018}^{\infty}\frac{\sin^3x}{x^{\frac{3}{4}}+x^{\frac{1}{6}}\cos x}\,dx$$

Now I thought about using Dirichlet's criteria, but unfortunately $\frac{1}{x^{\frac{3}{4}}+x^{\frac{1}{6}}\cos x}$ is not monotonically decreasing. No relevant change of variables comes to mind also. Am I on the right track?
Any help would be appreciated.
 A: Note that $\sin^3 x = \frac{3}{4} \sin x - \frac{1}{4} \sin 3x$ and it is enough to consider the convergence as $c \to \infty$ of the integral
$$\tag{*}\int_{2018}^c\frac{\sin ax}{x^{3/4} + x^{1/6}\cos x}\, dx = \int_{2018}^c\frac{\sin ax}{x^{3/4} }\, dx - \int_{2018}^c\left[\frac{\sin ax}{x^{3/4} + x^{1/6}\cos x} - \frac{\sin ax}{x^{3/4} }\right]\, dx$$
The first integral on the RHS of (*) converges by the Dirichlet test.  For the second integral we have
$$\left|\frac{\sin ax}{x^{3/4} + x^{1/6}\cos x} - \frac{\sin ax}{x^{3/4} }\right| = \left|\frac{\sin ax\cdot x^{1/6}\cos x}{(x^{3/4} + x^{1/6}\cos x)x^{3/4}} \right| = \frac{|\sin ax|\, |\cos x|}{(x^{3/4} + x^{1/6}\cos x)x^{7/12}}\\ \leqslant \frac{1}{(x^{3/4} + x^{1/6}\cos x)x^{7/12}} = \frac{1}{x^{4/3}+ x^{3/4}\cos x}$$
Consequently, the second integral on the RHS of (*) is absolutely convergent by the comparison test, since by the limit comparison test (with $x^{-4/3}$) we have
$$\int_{2018}^\infty \frac{dx}{x^{4/3}+ x^{3/4}\cos x} < \infty$$
Therefore, the improper  integral on the LHS of (*) is convergent.
