The theoretical tools at my disposal are Abel's test and Dirichlet's test. To recap those: Say I have an integral of the form $$\int_{a}^{b}f\cdot g \hspace{1.5mm} dx$$ with Improperness (vertical or horizontal asymptote) at b.
Abel's test guarantees convergence for
$\bullet$ $g$ monotone and bounded on $(a,b)$ $\hspace{5mm}$ $\bullet$ $\int_{a}^{b}f $ convergent.
Dirichlet's test guarantees convergence for
$\bullet$ $g$ monotone on $(a,b)$ and $lim_{x\to b}\hspace{1.5mm} g(x) = 0 $ $\hspace{5mm}$ $\bullet$ $\lim_{\beta \to b}$ $\int_{a}^{\beta}f $ bounded.
So for $\int_{1}^{\infty} \frac{cos(x)}{x^r} dx$, $cos(x)$ must be my $f$ and since $\int_{1}^{\infty} cosx \hspace{1mm} dx$ diverges, I must show it is bounded, which is intuitively easy given the graph of $cos(x). $ But how to make it precise? With that done, for $r > 0 $ I get my bounded function $ g(x)=\frac{1}{x^r} $ as required by Dirichlet's test, right?
Disclaimer: This is my first answer. And my English is very bad.
First, you know the inequality $$(\forall r \in \mathbb{R})~~~ \frac{\cos(x)}{x^r} \leq \frac{1}{x^r}$$
With integration by parts you can prove convergence for $r=1$ using $$\frac{\cos(x)}{x} \leq \frac{1}{x}$$
For $r > 1$ again use integration by parts: $\int_{1}^{+\infty} \frac{\cos(x)}{x^a} dx= \left[\frac{\sin(x)}{x^r}\right]_1^{+\infty} + r\int_1^{+\infty} \frac{\cos(x)}{x^{r+1}}dx \leq \cos (1)+ r\int_1^{+\infty} \frac{\cos(x)}{x^{r+1}} dx$ and
$$\int_1^{+\infty} \frac{|\cos(x)|}{x^{r+1}} dx\leq \int_1^{+\infty} \frac{1}{x^{r+1}} dx<\infty.$$ Hence your integral converges on $[1,+\infty) \Leftrightarrow r>1$.
Similar arguments show divergence for $r < 1$.