Is there a trick to solve $\int_{-1}^1 \frac{P(t)}{\sqrt{1-t^2}}{\rm d}t=a[P(x_1)+P(x_2)+P(x_3)]$?

Note that $\int_{-1}^{1}\frac{at^3+bt^2+ct+d}{\sqrt{1-t^2}}dt$ is, by symmetry, equal to $\int_{-1}^{1}\frac{bt^2+d}{\sqrt{1-t^2}}dt$. This evaluates to $\frac{\pi}{2}(b+2d)$. Now we set the first constant to $\frac{\pi}{2} k$ and try $x_1=0, x_2=z$ and $x_3=-z$, again to exploit the symmetry of the polynomial. The equality the becomes $\frac{\pi}{2}(b+2d) = \pi kbz^2+\frac{3}{2}\pi kd$. Comparing coefficients you should get $k=\frac{2}{3}$ and $z=\frac{\sqrt{3}}{2}$. So we have $\int_{-1}^{1}\frac{at^3+bt^2+ct+d}{\sqrt{1-t^2}}dt = \frac{\pi}{3}[P(-\frac{\sqrt{3}}{2})+P(0)+P(\frac{\sqrt{3}}{2})]$.

Surely this is not the most elegant solution but it works just fine for degree 3 or 5 polynomials.

This probably has something to do with Chebyshev polynomials. \begin{align} T_0(x)=&\ 1\\ T_1(x)=&\ x\\ T_{n+1}(x)=&\ 2xT_n(x)-T_{n-1}\hspace{1em} \forall\ n\geq 1 \end{align}

These polynomials have orthogonality property with $1/\sqrt{1-x^2}$ as weight. $$\int_{-1}^{1}T_n(x)T_m(x)\frac{1}{\sqrt{1-x^2}}\,\text{d}x= \left\lbrace \begin{array}{ll} 0 & m\neq n\\ \pi & m=n=0\\ \pi/2 & m=n\neq 0 \end{array} \right.$$

For example, in a third order polynomial, $1=T_0(x)T_0(x)$, $x=T_1(x)T_0(x)$, $x^2=T_1(x)T_1(x)$ and $x^3=0.5T_1(x)\times(T_2(x)+T_0(x))$.

P.S. I am not allowed to comment given my reputation and hence am posting this as an answer; although, this should have been a comment.

Edit

These polynomials also satisfy what is called "discrete orthogonality" (see this). If $x_i$ are the roots of $T_n(x)$, then for $0\leq p,q\leq n-1$, the following holds. \begin{align} \sum_i T_p(x_i)T_q(x_i)=\left\lbrace \begin{array}{ll} 0 & p\neq q\\ n & p=q=0\\ n/2 & p=q\neq 0\\ \end{array} \right. \end{align}

With this property, I think the question is solved. Let $P_n(x)$ be any $n$-th order polynomial. Then, $P_n(x)$ can be expanded using the Chebyshev polynomial basis. $$P_n(x)=\sum_{i=0}^{n} b_i T_i(x)=\sum_{i=0}^{n} b_i T_i(x)T_0(x)$$

By orthogonality, as already pointed out in the comments, $$\int_{-1}^{1} \frac{P_n(x)}{\sqrt{1-x^2}}=\int_{-1}^{1} \frac{P_n(x)T_0(x)}{\sqrt{1-x^2}}=\pi b_0$$

Now, suppose $x_j$, $0\leq j\leq n-1$ represent roots of $T_n(x)$. Then, $$\sum_{j=0}^{n-1}P_n(x_j)=\sum_{j=0}^{n-1}\sum_{i=0}^{n} b_i T_i(x_j)T_0(x_j)=\sum_{i=0}^{n} b_i \sum_{j=0}^{n-1} T_i(x_j)T_0(x_j)=nb_0$$

The last equality follows from discrete orthogonality. Summing up, we have $$\int_{-1}^{1} \frac{P_n(x)}{\sqrt{1-x^2}}=\pi b_0=(\pi/n) \sum_{j=0}^{n-1}P_n(x_j)$$