In answering a question on math.SE, I attempted to find integral of Fejér kernel by using
$ K_n(t) = \frac{1}{n} U_{n-1}^2\left( \cos \frac{t}{2} \right) $ where $U_n(z)$ stands for the Chebyshev polynomial of the second kind. Then $ \frac{1}{2 \pi} \int_{-\pi}^\pi K_n(t) \mathrm{d} t = \frac{1}{\pi n} \int_{-1}^1 \frac{U_{n-1}^2(z)}{\sqrt{1-z^2}} \mathrm{d} z $ Also it is known that the left-hand-side integral is one, I could not find any neat way of showing that the right-hand-side integral equals one.
Note that, by orthogonality property for $U_n(z)$: $ \int_{-1}^1 U_{n-1}^2(z) \sqrt{1-z^2} \mathrm{d} z = \frac{\pi}{2} $
Thanks for reading.