Gauss-Legendre quadrature rule

The integral we want to evaluate is:
$$E(x)=\int_0^{\pi/2}\sqrt{1-x^2\sin(\theta)^2}\,d\theta$$
for $x=\sqrt 3/2$. We write this as:
$$E(x) =(1+\frac\pi2) \int_{-1}^1 \sqrt{1-x^2\sin((1+\frac\pi2) \theta-1)^2}\,d\theta\tag{$*$}$$
The zeros of the Legendre Polynomial $P_5$ are given by (taken from Wikipedia):
$$x_1=0,\ x_{2,3}=\pm\frac13\sqrt{5-2\sqrt{10/7}}, \ x_{4,5}=\pm\frac13\sqrt{5+2\sqrt{10/7}}$$
with associated weights:
$$w_1=\frac{128}{225},\ w_{2,3}=\ \frac{322+13\sqrt{70}}{900}, \ w_{4,5}\frac{322-13\sqrt{70}}{900}$$
We evaluate the integrand $f$ of $(*)$ at the points $x_i$ (with the number $x=\sqrt{3}/2$), but these values dont really admit any simplifications. We calculate the endresult in Mathematica with the formula:
$$\sum_{i=1}^5 f(x_i)\,w_i\approx 1.234129947$$
Rounding to $6$ decimal places yields
$$1.23413$$
as an answer

See the attached document for the calculations in Mathematica.