Limit of integral of sequences 
Calculate
  $$\lim_{n\to\infty}\int_{-\pi/4}^{\pi/4}\frac{n\cos(x)}{n^2x^2+1}\,dx$$

I don't know how to calculate the integral and the sequence is not monotone or dominated by a $L^1$ function, so I'm stucked. Any idea?
 A: How about integration by parts?
\begin{align*}
{\int\frac{n\cos (x)}{n^2x^2+1}\mathrm{d}x}=\arctan(nx)\cos(x)+\int\arctan(nx)\sin(x)\,\mathrm{d}x.
\end{align*}
The absolute value of $\arctan(n x)\sin(x)$ is uniformly bounded above by the constant function $\pi/2$, which is integrable on $[-\pi/4,\pi/4]$, so you can use the dominated convergence theorem.
To wrap up,
\begin{align*}
&\,\lim_{n\to\infty}\int_{-\pi/4}^{\pi/4} \frac{n\cos (x)}{n^2 x^2+1}\,\mathrm{d}x=\lim_{n\to\infty}\left\{\left[\arctan(nx)\cos (x)\right]_{x=-\pi/4}^{x=\pi/4}\right\}+\lim_{n\to\infty}\int_{-\pi/4}^{\pi/4}\arctan(nx)\sin(x)\,\mathrm{d}x\\
=&\,\lim_{n\to\infty}\left\{\arctan\left(\frac{n\pi}{4}\right)\times\frac{1}{\sqrt{2}}-\arctan\left(-\frac{n\pi}{4}\right)\times\frac{1}{\sqrt{2}}\right\}+\int_{-\pi/4}^{\pi/4}\lim_{n\to\infty}\left[\arctan(nx)\sin(x)\right]\mathrm{d}x\\
=&\,\lim_{n\to\infty}\left\{2\arctan\left(\frac{n\pi}{4}\right)\times\frac{1}{\sqrt{2}}\right\}+\int_{-\pi/4}^0\left[\lim_{n\to\infty}\arctan(\underset{\color{red}-}{n x})\right]\sin(x)\,\mathrm{d}x\\
+&\,\int_{0}^{\pi/4}\left[\lim_{n\to\infty}\arctan(\underset{\color{red}+}{n x})\right]\sin(x)\,\mathrm{d}x\\
=&\,2\times\frac{\pi}{2}\times\frac{1}{\sqrt{2}}+\int_{-\pi/4}^0\left(-\frac{\pi}{2}\right)\times\sin(x)\,\mathrm{d}x+\int_{0}^{\pi/4}\frac{\pi}{2}\times\sin(x)\,\mathrm{d}x\\
=&\,\frac{\pi}{\sqrt{2}}+\left(-\frac{\pi}{2}\right)\times\left(-1+\frac{1}{\sqrt{2}}\right)+\left(\frac{\pi}{2}\right)\times\left(1-\frac{1}{\sqrt{2}}\right)=\pi.
\end{align*}
A: 
Use the fact that $1-x^2\leqslant\cos x\leqslant1$ for every $x$, and the rest follows easily.

To wit, the $n$th integral $I_n$ is such that $J_n-K_n\leqslant I_n\leqslant J_n$ with
$$
J_n=\int_{-\pi/4}^{\pi/4}\frac{n\mathrm dx}{n^2x^2+1},\qquad K_n=\int_{-\pi/4}^{\pi/4}\frac{nx^2\mathrm dx}{n^2x^2+1}.
$$
The change of variable $s=nx$ yields
$$
J_n=\int_{-n\pi/4}^{n\pi/4}\frac{\mathrm ds}{s^2+1}\to\int_{-\infty}^\infty\frac{\mathrm ds}{s^2+1}=J,
$$
and
$$
n^2K_n=\int_{-n\pi/4}^{n\pi/4}\frac{s^2\mathrm ds}{s^2+1}\leqslant\int_{-n\pi/4}^{n\pi/4}\mathrm ds=\frac{n\pi}2.
$$
Thus, $K_n\to0$ and $I_n\to J$ where, of course,
$$
J=\left.\arctan s\right|_{-\infty}^\infty=\pi.
$$
