My attempt: $$\int_{1}^{\infty}\frac{\cos(x) dx}{1+x^2} \le \int_{1}^{\infty} \frac{dx}{1+x^2}$$ and the latter integral converges, so $\int_{1}^{\infty}\frac{\cos(x) dx}{1+x^2}$ also converges. And similar argument follows for $\int_{1}^{\infty}\frac{\sin(x) dx}{1+x^2}$.
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Is this correct ? Then, is it true that all integrals bounded from above (strictly or non-strictly) by a finite value converge?
As mentioned in the comments, your solution is wrong. Comparison tests work only for non-negative (or similarly non positive) functions. You cannot use this for functions like $\sin$ or $\cos$. However, you can say that $\frac{|\cos(x)|}{1+x^2}\leq\frac{1}{1+x^2}$, and so by the comparison test $\int_1^{\infty}\frac{|\cos(x)|}{1+x^2}dx<\infty$. Thus the integral $\int_1^{\infty}\frac{\cos(x)}{1+x^2}dx$ converges absolutely, and so converges.
You can also prove the convergence of the original integral directly if you are familiar with Dirichlet's test. But that solution would be kind of a waste, as proving absolute convergence is proving a stronger statement. (and it's also easier)