I have problems proving the following statement:
$$\int_{0}^{\frac{2 \pi}{n}}\left|\cos \left(\frac{n}{2} x\right)\right| d x=\int_{0}^{\frac{2 \pi}{n}} \left| \sin \left(\frac{n}{2} x\right)\right| d x=\frac{4}{n}$$
The second equality is clear to me because it is straightforward integration: $$\left.\int_{0}^{\frac{2 \pi}{n}}\left|\sin \left(\frac{n}{2} x\right)\right| d x=\int_{0}^{\frac{2 \pi}{n}} \sin \left(\frac{n}{2} x\right) d x=\frac{-\cos \left(\frac{n}{2} x\right)}{\frac{n}{2}}\right]_{0}^{\frac{2 \pi}{n}}=\frac{4}{n}$$
However, I don't know how to prove it analytically with $\displaystyle \left|\cos \left(\frac{n}{2} x\right)\right|$ because the absolute value changes in the integration domain.