Prove $\int\limits_0^{\frac{\pi}{2}}\sin(2n x)\,\cot x\,\mathrm{d}x- \int\limits_0^{\frac{\pi}{2}}\frac{\sin(2n x)}{x}\,\mathrm{d}x\to 0$ 
Let $$a_n=\int\limits_0^{\frac{\pi}{2}}\sin(2n x)\,\cot x\,\mathrm{d}x ~~~\textrm{ and} ~~~   b_n=\int\limits_0^{\frac{\pi}{2}}\frac{\sin(2n x)}{x}\,\mathrm{d}x.$$
  Prove that $a_n-b_n \to 0.$

Attempt. We have that 
$$a_n-b_n  = \int\limits_0^{\frac{\pi}{2}}\sin(2n x)\,\left(\cot x-\frac{1}{x}\right)\,\mathrm{d}x.$$
Integration by parts would give:
$$a_n-b_n=\left[\log\left(\frac{\sin x}{x}\right)\sin(2nx)\right]_0^{\frac{\pi}{2}}-2n\int\limits_0^{\frac{\pi}{2}}\cos(2n x)\,\log\left(\frac{\sin x}{x}\right)\,\mathrm{d}x$$
$$=-2n\int\limits_0^{\frac{\pi}{2}}\cos(2n x)\,\log\left(\frac{\sin x}{x}\right)\,\mathrm{d}x$$ but this doesn't seem to go any further.
On the other hand, from Integral $\int_0^\pi \cot(x/2)\sin(nx)\,dx$ we have $a_n=\frac{\pi}{2}$, so:
$$b_n-a_n=\int\limits_0^{\frac{\pi}{2}}\left(\frac{\sin(2n x)}{x}-1\right)\,\mathrm{d}x,$$
so:
$$|b_n-a_n|\leqslant \int\limits_0^{\frac{\pi}{2}}\left|\frac{\sin(2n x)}{x}-1\right|\,\mathrm{d}x,$$
but I also didn't manage to get this any further.
Thanks in advance for the help.

Edit. Consequence of the above limit is the evaluation of the Dirichlet integral: $$\int\limits_0^{+\infty}\frac{\sin x}{x}\,\mathrm{d}x=
\lim_{n \to +\infty}b_n=\lim_{n \to +\infty}a_n=\frac{\pi}{2}.$$
 A: N.B.: I think it's good, but I'm tired, so it might be wrong. I'm going to check it after I wake up; read it carefully until then.
Let $f(x)=\cot(x)-\frac{1}{x}$ with $f(0)=0$.
We have that
\begin{align}
\left|\int_0^{\pi/2}f(x)\sin(2nx)\mathrm{d}x\right|
&=\left|-\int_0^{\pi/2}f'(x)\frac{\cos(2nx)}{2n}\mathrm{d}x+\left[\frac{f(x)\cos(2nx)}{2n}\right]_0^{\pi/2}\right|\\
&=\left|-\int_0^{\pi/2}f'(x)\frac{\cos(2nx)}{2n}\mathrm{d}x+\left(-\frac{2}{\pi}\right)\frac{\cos(n \pi)}{2n}\right|\\
&=\left|-\int_0^{\pi/2}f'(x)\frac{\cos(2nx)}{2n}\mathrm{d}x-\frac{(-1)^n}{n \pi}\right|\\
&\leqslant \left|\int_0^{\pi/2}f'(x)\frac{\cos(2nx)}{2n}\mathrm{d}x\right|+\left|\frac{(-1)^n}{n\pi}\right| \\
&\leqslant \int_0^{\pi/2}\left|f'(x)\frac{\cos(2nx)}{2n}\right|\mathrm{d}x+\frac{1}{n\pi}\\
&\leqslant \int_0^{\pi/2}\left|C\frac{\cos(2nx)}{2n}\right|\mathrm{d}x+\frac{1}{n\pi}\\
&\leqslant \frac{C}{2n}\int_0^{\pi/2}\left|\cos(2nx)\right|\mathrm{d}x+\frac{1}{n\pi}\\
&\leqslant \frac{C}{2n}\int_0^{\pi/2}1\mathrm{d}x+\frac{1}{n\pi}\\
&\leqslant \frac{\pi C}{4n}+\frac{1}{n\pi} \to 0
\end{align}
Since the absolute value of the derivative of $f$ on $[0, \pi/2]$ is bounded by a constant $C$.
