Show if the series $\sum^{\infty}_1\sin\frac{1}{n}$ converges using the integral test I have to examine if the series $$\sum^{\infty}_1\sin\frac{1}{n}$$ converges using the integral test.
The series is given by function $f(x)=\sin\frac{1}{x}$ which is positive when $x\ge 1$. The series can be written as an improper integral $$\sum^\infty_1\sin\frac{1}{x}=\int^\infty_1 \sin\frac{1}{x}dx$$
and if the improper integral converges, the series converges by the integral test.
With substitution $u=\frac{1}{x}$ and $du=-\frac{1}{x^2}dx$ where the new bounds are $\frac{1}{1}=1$ and $\frac{1}{\infty}=0$,
$$\int^\infty_1 \sin\frac{1}{x}dx=\int^\infty_1-x^2\sin udu.$$
However, I'm not sure how to go on from here due to the $-x^2$.

First, it's not that $\sum_{n=1}^\infty \sin\frac{1}{n}=\int_{1}^\infty\sin\frac{1}{x}dx$, it is that the two converge or diverge simultaneously. We can provide an upper and lower bound for the integral in terms of the sum plus some constants, be they will differ in general.
As for your $x^2$, we set $u=\frac{1}{x}$, so $x=\frac{1}{u}$, so you are interested in the convergence of the integral
$$\int_0^1 \frac{1}{u^2}\sin(u)du$$ which diverges.
To see that this diverges, we use the inequality $\sin(u)\geq u-\frac{u^3}{6}$ for $u$ small. Then since $\sin(u)\geq 0$ for $u\in[0,1]$, we have
$$\int_0^1\frac{1}{u^2}\sin(u)du\geq \int_0^\epsilon\frac{1}{u^2}\sin(u)du\geq \int_0^\epsilon \frac{1}{u}-\frac{u}{6}du$$
for some small epsilon.This clearly diverges.
