Integral of $\left(\sum\limits_n\cos\theta_n\right)^m\prod\limits_{r<s}(\cos\theta_r-\cos\theta_s)^2\prod\limits_j(1+\cos\theta_j)$ on $[0,\pi]^k$

Prove that for positive integers $k$, $m$ $$\frac{{2^k}^{2-2k}}{\pi^k k !}\int_{[0,\pi]^k} (2\cos\theta_1 + \cdots +2\cos\theta_k)^m \prod_{1\leq r <s \leq k}(\cos\theta_r - \cos\theta_s)^2 \prod\limits_{i=1}^k (1+\cos\theta_i)\,d\theta_1 \ldots d\theta_k\\ = \sum\limits_{t_1+ \cdots +t_k=m}{m \choose t_1,\ldots,t_k} \det {\left({{t_i+2k-i-j}\choose {\lfloor {\frac {t_i+2k-i-j}{2}}\rfloor }}\right)}^k_{i,j=1}$$ I have no idea how to begin to solve this question. It would be easy for me if you give me some hints or some steps to solve this.

Hint : the first product is the square of a Vandermonde determinant, let's name it $V(\theta)^2$. If you develop the term in $(2 \cos(\theta_1)+...+ 2 \cos(\theta_k))^m$, you get by the multinomial formula

$$\sum_{t_1+...+t_k = m} {m \choose t_1, ..., t_k} \int (2\cos(\theta_1))^{t_1}...(2\cos(\theta_{k}))^{t_k} V(\theta)^2 d\mu(\theta_1)...d\mu(\theta_k)$$

with $\mu$ the measure whose density wrt the Lebesgue measure on $[0,\pi]$ is $1+\cos(\theta)$.

Now, there is this famous Andreief's formula (basically, it's just a generalization of Fubini-Tonnelli) which states that if $f_1, ..., f_k, g_1, ..., g_k$ are in $L^2(\mu)$, then

$$\int...\int \mathrm{det}(f_i(x_j))\mathrm{det}(g_i(x_j))\prod_{i=1}^k d\mu (x_i) = n!\cdot \mathrm{det}\left( \int...\int f_i(x) g_j(x) d\mu(x) \right)$$

(this is true for every measure $\mu$)

You want to apply this to the integrals $\int (2\cos(\theta_1))^{t_1}...(2\cos(\theta_{k}))^{t_k} V(\theta)^2 d\mu(\theta_1)...d\mu(\theta_k)$, by inserting the cosines into the corresponding rows of the Vandermonde determinant.