Determine if $\sum^{\infty}_{n=1}\int^{\frac{\sin n}{n}}_0\frac{\sin x}{x} \, dx $ is converges or diverges.

This is edited version. Thanks to Tigran Saluev for pointing out the mistatke.

By his argument your series converges iff the series $$ \sum\limits_{n=1}^\infty\frac{|\sin n|}{n}\tag{1} $$ converges. Note that $$ \sum\limits_{n=1}^\infty\frac{|\sin n|}{n}\geq\sum\limits_{n=1}^\infty\frac{\sin^2 n}{n}=\frac{1}{2}\sum\limits_{n=1}^\infty\frac{1}{n}-\frac{1}{2}\sum\limits_{n=1}^\infty\frac{\cos 2n}{n} $$ The first term diverges as harmonic series diverges. The second term converges since $$ \sum\limits_{n=1}^\infty\frac{\cos 2n}{n} =\Re\left(\sum\limits_{n=1}^\infty\frac{e^{i\cdot 2n}}{n}\right) =\Re\left(\sum\limits_{n=1}^\infty\frac{z^n}{n}\biggl|_{z=e^{2i}}\right) =\Re\left(-\ln(1-z)|_{z=e^{2i}}\right)<\infty $$ Thus the series $(1)$ diverges, hence your series diverges too.

$\frac{\sin x}{x}$ is a continuous function, thus $$\lim_{a\rightarrow 0}\int_0^a \frac{\sin x}{x}\,dx\ \ \ \mbox{ is always equal to } 0.$$ But that is not enough.

$\frac{\sin x}{x}$ is also positive in a neighborhood of 0, thus all the integrals for $n$ greater than some fixed $N$ are positive. It doesn't depend on the sign of $\frac{\sin n}{n}$, since the integrated function is even. But the fact $\frac{\sin n}{n}$ is not necessarily positive forbids you to write $$\int_0^{\frac{\sin n}{n}} \frac{\sin x}{x}\,dx \le \frac{\sin n}{n} $$ like Norbert has done. Correct estimate looks like $$\int_0^{\frac{\sin n}{n}} \frac{\sin x}{x}\,dx \le \frac{|\sin n|}{n}.$$ So the solution is not that simple.

$\frac{\sin x}{x}$ is decreasing and convex on $[0, 1]$:

So we can estimate the integral from below as the square of highlighted region: $$ \int_0^{\frac{\sin n}{n}} \frac{\sin x}{x}\,dx \ge \left.\frac{\sin x}{x}\right|_{x=\frac{|\sin n|}{n}}\cdot\frac{|\sin n|}{n} + \frac{1}{2}\cdot\frac{|\sin n|}{n}\cdot\left(1 - \left.\frac{\sin x}{x}\right|_{x = \frac{\sin n}{n}}\right) = $$ $$ = \frac{1}{2}\sin\left(\frac{|\sin n|}{n}\right) + \frac{1}{2}\frac{|\sin n|}{n}. $$ Thus, including my remark on Norbert's answer, your series converges if and only if $$\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$$ converges.