Determine if the following integral absolutly converges, conditionaly converges or diverges.
$$ \int_{2018}^{\infty} \frac{\cos x}{x + x^{1/3}\sin^2x}\mathrm dx $$
I thought maybe to try dirichle:
$$ f(x) = x + x^{1/3}\sin^2x, g(x) = \cos x $$
But $f(x)$ is not monotonic
I thought to look at the absolute value:
$$ \left|\frac{\cos x}{x + x^{1/3}\sin^2x}\right|\le \frac{1}{|x + x^{1/3}\sin^2x|} $$
But i dont see where to go from here.
I dont have an idea, stuck, can you help?
Use integration by parts with $dv=\cos(x)\,dx$: $$ \int_{2018}^{\infty} \frac{\cos (x)}{x + x^{1/3}\sin^2x}\,dx $$ $$ = \left. \frac{\sin (x)}{x + x^{1/3}\sin^2x}\right|_{2018}^{\infty}+\int_{2018}^{\infty} \frac{\frac{1}{3}\sin^2(x) x^{-2/3} + x^{1/3} \sin(2x) + 1}{(x + x^{1/3} \sin^2(x))^2}\,dx $$The boundary term is finite and the integrand in the new integral is $O(x^{-5/3})$, which is improperly integrable. So the integral converges, but not absolutely. To see this, note: $$ \int_{2018}^{\infty} \left|\frac{\cos (x)}{x + x^{1/3}\sin^2x}\right|\,dx $$ $$ >\int_{2018}^{\infty} \left|\frac{\cos (x)}{10x}\right|\,dx $$ $$ >\frac{1}{10}\int_{643\pi}^{\infty} \left|\frac{\cos (x)}{x}\right|\,dx $$ $$ =\frac{1}{10}\sum_{n=643}^{\infty}\int_{n\pi}^{(n+1)\pi} \left|\frac{\cos (x)}{x}\right|\,dx $$ $$ >\frac{1}{10}\sum_{n=643}^{\infty} \frac{1}{n\pi}\int_{n\pi}^{(n+1)\pi} \left|{\cos (x)}\right|\,dx $$ $$ =\frac{1}{5\pi}\sum_{n=643}^{\infty} \frac{1}{n}=\infty $$