Is there a neat way to show $\int_{-1}^1 \frac{ U_n(z) U_n(z)}{\sqrt{1-z^2}} \mathrm{d} z = \pi (n+1)$ 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.
 A: The weight $\frac{1}{\sqrt{1-z^2}}$ may not be the one used for the orthogonality relations of $U_n(x)$, but it is however the weight used for the orthogonality relations of $T_n(x)$.  Specifically, we have that for $n,m>0$ $$\int_{-1}^1 T_n(z)T_m(z)\frac{dz}{\sqrt{1-z^2}}=\frac{\pi}{2}\delta_{n,m}.$$ (in the case that $m=n=0$, then the integral equals $\pi$) 
Hence we need only find a way to relate $U_n(x)$ to a sum of $T_n(x)$ terms.  On this note, we have the identities:
$$n\ \text{odd}\Rightarrow\ \ U_n(x)= 2\sum_{k\leq n,\ k\ \text{odd}} T_k (x)$$
and
$$n\ \text{even}\Rightarrow\ \ U_n(x)= -1+2\sum_{k\leq n,\ k\ \text{even}} T_k (x).$$
These both follow from induction and the recurrence relations for $U_n(x),\ T_n(x)$.
If $n$ is odd, then by the above we see that the integral will give $\frac{\pi}{2}$ for each diagonal term, and zero on all the others.  Hence we conclude that $$\int_{-1}^1 \frac{U_n(z)U_n(z)}{\sqrt{1-z^2}}dz =  4\sum_{k\leq n,\ k\ \text{odd}} \frac{\pi}{2}= \pi (n+1).$$
If $n$ is even, things follow in a similar manner.
A: A much simpler approach is to let $x=\cos\theta.$ Then $$\int_{-1}^{1}\frac{U_{n-1}(x)^{2}}{\sqrt{1-x^{2}}}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}U_{n-1}(\cos\theta)^{2}d\theta.$$ Since $U_{n-1}(\cos\theta)=\frac{\sin(n\theta)}{\sin\theta},$ our integral becomes $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(n\theta)^{2}}{\sin(\theta)^{2}}d\theta,$$ and for $n=2k+1,$ we can use the identity $$\frac{\sin((2k+1)\theta)}{\sin(\theta)}=1+\sum_{j=1}^{k}\cos(2jx)$$ to evaluate the integral, and similarly for $k$ even.
