I am wondering whether there is a known closed form for the $k$th Chebyshev coefficient of a $n$th Chebyshev interpolant, that is, $$\int_{-1}^1T_k(x)L_{n,i}(x)\frac{dx}{\sqrt{1-x^2}}$$ where $T_k$ is the $k$th Chebyshev polynomial (of the 1st kind), $t_{n,j}:=-\cos\frac{2\pi j}{n}$ for $0\le j\le n$ is a Chebyshev node (also the roots of $T_n$), $L_{n,i}(x):=\prod\limits_{j\neq i}\frac{x-t_{n,j}}{t_{n,i}-t_{n,j}}$, and $k<n$ (as otherwise the integral is trivially 0, from $\deg L_{n,i}=n-1$). Notice that $L_{n,i}(x)\propto\frac{T_n(x)}{x-t_{n,i}}$, so evaluating the desired integral is equivalent to evaluating $$\int_{-1}^1\frac{T_k(x)T_n(x)}{x-t_{n,i}}\frac{dx}{\sqrt{1-x^2}}.$$
The only idea I have for this is is that $2T_kT_n=T_{n+k}+T_{n-k}$ and $2xT_n(x)=T_{n-1}(x)+T_{n+1}(x)$; from the latter, we can find a closed form for $x^mT_n(x)$ as a $\mathbb{Q}$-linear sum of Chebyshev polynomials (I think the coefficients will be something like binomial coefficients times powers of 2 -- nothing too crazysee edit 2), so using $\frac{1}{1-x}=\sum x^m$ we get that the integral becomes $$\int_{-1}^1\frac{T_k(x)T_n(x)}{x-t_{n,i}}\frac{dx}{\sqrt{1-x^2}}=-\frac{1}{2t_{n,i}}\int_{-1}^1\sum_m(T_{n+k}(x)+T_{n-k}(x))\left(\frac{x}{t_{n,i}}\right)^m\frac{dx}{\sqrt{1-x^2}}$$
and then formulaically computing the relevant coefficients, however resolving this seems relatively involved.
EDIT: This approach will workis promising because we only need to track the coefficient for each $m$ of $T_0\equiv1$ by orthogonality $(\int_{-1}^1T_j(x)\frac{dx}{\sqrt{1-x^2}}=0$ for $j>0$). However it involves looking at the power series outside of its radius of convergence. So I guess what I'm asking is if there is a "more intrinsic"/less-computationally-involved way to come up with a value of either of the first two integrals (above the horizontal line).
EDIT 2: It turns out that $$\int_{-1}^1x^mT_n(x)\frac{dx}{\sqrt{1-x^2}}=\frac{1}{2^m\pi}\binom{m}{\frac{m-n}{2}},$$ defined to be 0 if $\frac{m-n}{2}$ isn't a nonnegative integer, the idea being that the recursion works out perfectly if we let $T_{-n}:=T_n$ so then $(2x)^mT_n(x)=\sum\limits_{j=0}^m\binom{m}{j}T_{n+m-2j}(x)$.