I doubt this is the best way, but it will do.
Firstly, notice that you have no problems on the interval $[0,1]$, and in fact the integral is bounded above by $1$ there. So we are left to consider
$$ \int_1^\infty x\cos(x^3)dx.$$
Perform the substitution $u = x^3$ to see that this is equivalent to the convergence of
$$ \int_1^\infty \frac{\cos(x)}{\sqrt[3] x} dx.$$
The numerator $\cos x$ oscillates positive and negative in perfectly regular sequences, positive in intervals of the form $x \in [2n\pi + 3\pi/2, 2n\pi + 5\pi/2]$, and negative in intervals of the form $x \in [2n\pi + \pi/2, 2n\pi + 3\pi/2]$. The denominator is always positive, and the fractions $\frac{1}{\sqrt[3] x}$ are always decreasing. So the area in each positive hump and each negative hump is decreasing, and going to zero.
As the humps alternate between positive and negative signs, and the areas in each hump are monotonically decreasing, the integral will converge to some limit $L$. This is a mimicry of the proof of the alternating series test.
In fact, if you call $H_{n}$ the area of the $n$th hump, so that $H_{2n} > 0$ and $H_{2n + 1} < 0$, then as $H_n$ is alternating, $\lvert H_{n + 1}\rvert < \lvert H_n \rvert$, and $\lvert H_n \rvert \to 0$, then
$$ \int_1^\infty \frac{\cos x}{\sqrt[3]x} dx = \sum_{n \geq 1} H_n = L$$
exactly by the classical alternating series test.