Proving $\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx=0 $

$$\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx=0 $$

I consider if it can be solved by other methods without Riemann lemma. I try my best to do it as follow:

Let $t=nx$ $$\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx$$ \begin{eqnarray} &=&\lim_{n\to \infty}\int_0^{n\pi}\frac{\sin\left(t\right)}{1+(\frac{t}{n})^2}d\frac{t}{n}\\ &=&\lim_{n\to \infty}\int_0^{n\pi}\frac{n\sin\left(t\right)}{n^2+t^2}dt\\ &=&\lim_{n\to \infty}\sum_{k=0}^{k=n-1}\int_{k\pi}^{(k+1)\pi}\frac{n\sin\left(t\right)}{n^2+t^2}dt\\ \end{eqnarray}

Then I can not go on. Who can tell me how to prove it? Thank you.

Integration by parts is enough:

$$\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx =\left.\frac{\frac{1}{n}(1-\cos(nx))}{1+x^2}\right|_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\frac{2x\,(1-\cos(nx))}{(1+x^2)^2}\,dx,\tag{1}$$ so: $$\left|\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx\right|\leq\frac{1}{n}\left(\frac{2}{\pi^2+1}+\frac{2\pi^2}{\pi^2+1}\right)=\frac{2}{n}.\tag{2}$$ Using the Cauchy-Schwarz inequality to bound the second integral in $(1)$, it is possible to improve the last inequality up to: $$\left|\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx\right|\leq\frac{1}{n}\left(\frac{2}{\pi^2+1}+\sqrt{\frac{3\pi}{2}\int_{0}^{\pi}\frac{4x^2}{(1+x^2)^4}\,dx}\right)<\frac{14}{9n}.\tag{3}$$

Consider your integral over two successive half-periods $$\int_{t=k\pi}^{k\pi+\pi}\frac{n\sin t}{n^2+t^2}dt+\int_{t=k\pi+\pi}^{k\pi+2\pi}\frac{n\sin t}{n^2+t^2}dt =\int_{t=k\pi}^{k\pi+\pi}\left(\frac{n\sin t}{n^2+t^2}-\frac{n\sin t}{n^2+(t+\pi)^2}\right)dt\\ =\int_{t=k\pi}^{k\pi+\pi}\frac{n(2\pi t+\pi^2)\sin t}{(n^2+t^2)(n^2+(t+\pi)^2)}dt=O(\frac1{n^2}).$$ The whole integral is $O(\frac1n)$.