Alternate proof of Dirichlet integral $\frac{\sin(x)}{x}$.

You can start by integrating the trigonometric form of the Dirichlet kernel.

$$\int_{-\pi}^{\pi}D_N(x)dx=2\pi=\int_{-\pi}^{\pi}\frac{\sin((N+1/2)x)}{\sin(x/2)}dx$$

Rewrite the integrand to make use of the second fact from the prompt.

$$\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{1}{\sin(x/2)}-\frac{2}{x}+\frac{2}{x})dx=2\pi$$

Split up the integral and consider the limit as $N\to\infty$.

$$\lim_{N\to\infty}\left[\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{1}{\sin(x/2)}-\frac{2}{x})dx+\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{2}{x})dx\right]=2\pi$$

The first integral vanishes in the limit per the Riemann-Lebesgue lemma. The second has an even integrand, so take the part from zero to $\pi$ and simplify.

$$\lim_{N\to\infty}\int_{0}^{\pi}\sin((N+1/2)x)(\frac{1}{x})dx=\frac{\pi}{2}$$

Finally, change variables to $y=(N+1/2)x$.

$$\lim_{N\to\infty}\int_{0}^{(N+1/2)\pi}\frac{\sin(y)}{y}dy=\frac{\pi}{2}$$