Since the partial sums of $\sum_{n}\sin(n)$ are bounded and $\sin(\frac{\pi}{2n})$ is a decreasing sequence that goes to $0$, Dirichlet test proves the convergence of your series.
To get a bound on the partial sums of $\sum_{n}\sin(n)$, note that $$ \left|\sum_{k=0}^n \sin(k)\right|=\left|\Im\sum_{k=0}^ne^{ik}\right|\leq\left|\frac{1-e^{i(n+1)}}{1-e^i}\right|\leq \frac{2}{|1-e^i|}$$
For $\beta>0$,
$$
\int_1^{+\infty}x^\alpha\sin(x^\beta)\mathrm dx=\frac{1}{\beta}\int_1^{+\infty}u^{\frac{\alpha + 1 - \beta}{\beta}}\sin{u}\mathrm du=\frac{1}{\beta}\int_1^{+\infty}\frac{\sin u}{u^{p}} \mathrm du
$$
with $p=1-\frac{\alpha + 1 }{\beta}$.
If $p >1$ the integral $\int_1^{+\infty}\frac{\sin u}{u^{p}}\mathrm du$ converges absolutely; in fact $\left|\frac{\sin u}{u^{p}}\right|\le\frac{1}{u^{p}}$ and $\int_1^{+\infty}\frac{1}{u^{p}}\mathrm du$ converges for $p>1$. From the Comparison Test, $\int_1^{+\infty}\left|\frac{\sin u}{u^{p}}\right|\mathrm du$ converges.
If $0< p\le 1$, the integral $\int_1^{+\infty}\frac{\sin u}{u^{p}}\mathrm du$ converges conditionally. For given $z\ge 1$, $$\left|\int_1^{z}\sin u\mathrm du\right|=\left|\cos 1-\cos z\right|\le 2$$ and $\frac{1}{u^{p}}$ approaches zero monotonically as $u\to\infty$ for $p>0$. Dirichlet's Test implies that $\int_1^{+\infty}\frac{\sin u}{u^{p}}\mathrm du$ converges for $p >0$. On the other hand, it is clear that
$$
\left|\frac{\sin u}{u^{p}}\right|\ge \frac{\sin^2 u}{u}=\frac{1}{2u}-\frac{\cos(2u)}{2u}
$$
where $\int_1^{+\infty}\frac{\cos(2u)}{2u}\mathrm du$ converges by Dirichlet's Test, and $\int_1^{+\infty}\frac{1}{2u}\mathrm du$ diverges. Hence for $0< p\le 1$, the integral $\int_1^{+\infty}\frac{\sin u}{u^{p}}\mathrm du$ converges conditionally.
Thus for $\beta>0$,
$$
\int_1^{+\infty}x^\alpha\sin(x^\beta)\mathrm dx
$$
converges absolutely for $\alpha<-1$ and conditionally for $-1\le\alpha<\beta-1$
In a similar way you can discuss the convergence of the integral for $\beta<0$,
$$
\int_1^{+\infty}x^\alpha\sin(x^\beta)\mathrm dx=\frac{-1}{\beta}\int_0^1 u^{\frac{\alpha + 1 - \beta}{\beta}}\sin{u}\mathrm du=\frac{-1}{\beta}\int_0^1\frac{\sin u}{u^{p}} \mathrm du=\frac{1}{\gamma}\int_0^1\frac{\sin u}{u^{p}} \mathrm du
$$
with $\gamma=-\beta>0$ and $p=1-\frac{\alpha + 1 }{\beta}=1+\frac{\alpha + 1 }{\gamma}$.
For $u\to 0$, we have $\sin u\sim u$ and then $\frac{\sin u}{u^{p}}\sim \frac{1}{u^{p-1}}$. Thus the integral converges for $p-1<1$, that is $p<2$ or
$$ \alpha+1 <\gamma=-\beta$$
Thus for $\beta<0$,
$$
\int_1^{+\infty}x^\alpha\sin(x^\beta)\mathrm dx
$$
converges for $ \alpha <-\beta-1$.
Best Answer
As pointed out by David Peterson in the comments, the error you make is splitting $$\frac{1}{2}\int_1^{\infty}1-\cos\left(\frac{6}{x}\right)\ dx= \int _1^{\infty }\frac{1}{2}dx - \frac{1}{2}\int _1^{\infty }\cos\left(\frac{6}{x}\right)\ dx$$ which you can only do when both integrals on the RHS converge.
In fact, the original improper integral is convergent. Since $$0\leq \sin^2\left(\frac{3}{x}\right)\leq \frac{9}{x^2}$$ and $$\int_1^{\infty}\frac{9}{x^2}\ dx$$ converges so does $$\int_1^{\infty} \sin^2\left(\frac{3}{x}\right) \ dx$$ by the comparison test. In fact $$\int_1^{\infty} \sin^{\alpha}\left(\frac{1}{x}\right)\ dx$$ converges for every $\alpha>1.$