Prove (using properties of definite integral) Prove the following:
$$\lim_{n \to \infty} \displaystyle \int_0 ^{2\pi} \frac{\sin nx}{x^2 + n^2} dx = 0$$
How would I prove this? I know you have to show your steps, but I'm literally stuck on the first one, so I can't. 
 A: The simplest approach seems to be to note that
$$
\left|\frac{\sin(nx)}{x^2+n^2}\right|\le\frac1{n^2}
$$
so that
$$
\left|\int_0^{2\pi}\frac{\sin(nx)}{x^2+n^2}\,\mathrm{d}x\right|\le\frac{2\pi}{n^2}
$$
A: HINT: Make use of 
$$\left|\dfrac{\sin nx}{x^2 + n^2}\right|\le\dfrac{1}{x^2 + n^2}$$
I think you also want to see Absolute value integral inequality proof step. 
A: We can come up with some bounds for this integral $I$. First notice that
$$|I|=\left|\int_0^{2\pi} \frac{\sin (nx)}{x^2+n^2} \operatorname{d}\!x \right| \le \int_0^{2\pi} \left|\frac{\sin (nx)}{x^2+n^2}\right| \operatorname{d}\!x \le \int_0^{2\pi} \frac{1}{\left|x^2+n^2\right|} \operatorname{d}\!x \le \int_0^{2\pi} \frac{1}{||x^2|-|n^2||} \operatorname{d}\!x$$
The last step was an application of the triangle inequaility: $|z+w| \ge ||z|-|w||$. Hence:
$$\lim_{n\to\infty}|I|\le\lim_{n\to\infty}\left(\int_0^{2\pi} \frac{1}{||x^2|-|n^2||} \operatorname{d}\!x\right) =\int_0^{2\pi}\lim_{n\to\infty}\left( \frac{1}{||x^2|-|n^2||}\right) \operatorname{d}\!x = 0$$
Since $|I|\to 0$ as $n \to \infty$ and so $I \to 0$ as $n \to \infty$.
