I am trying to show that the integral $\int_{-2}^2 U_n \left (\frac{x}{2}\right) \frac{p+1}{\pi}\frac{\sqrt{1-\frac{x^2}{4}}}{\left ( \sqrt{p}+\frac{1}{\sqrt{p}} \right )^2 - x^2} dx$ equals $p^{-n/2}$, when $n$ is an even integer. Here, $U_n(x)$ is the $n$-th Chebyshev polynomial of the second kind, defined as $U_n(x) = \frac{sin((n+1)cos^{-1}x)}{\sqrt{1-x^2}}$; $p$ is a prime, $n \geq 0$.
The corresponding integral when $n$ is odd is zero as the integrand is an odd function. However, I am not able to solve this integral for the even case. The integrand simplifies a bit once the definition of $U_n$ is used, but I am not able to deal with the $ \left ( \sqrt{p}+\frac{1}{\sqrt{p}} \right )^2 - x^2$ term. Any help?
The context is in analytic number theory, particularly results on vertical Sato-Tate. The measure here is actually $\mu_p(x) = \frac{p+1}{\pi}\frac{\sqrt{1-x^2}}{\left ( \sqrt{p}+\frac{1}{\sqrt{p}} \right )^2 - x^2} dx$, which is the Plancherel measure on GL$_2(\mathbb{Q}_p)$. I don't think this should be relevant to the integral though, I think the integral can be solved by simple calculus techniques.