These four integrals all equal …: …\int_0^1 \frac{dx}{\sqrt{x(1-x)}}\int_{-\infty}^\infty \frac{\sin x}{x}dx\frac{8}{3}\int_0^\pi \sin^4 x…

These four integrals all equal $\pi$:
$${\int }_0^1\frac{dx}{\sqrt{x(1-x)}}{\int }_{-\infty }^{\infty }\frac{\sin x}{x}dx\frac{8}{3}{\int }_0^{\pi }{\sin }^{4}xdx{\int }_0^{\infty }\frac{x^{-1/2}dx}{1+x}$$
(a) Apply the midpoint rule to two of them until $\pi \approx 3.1416$.
(b) Optional: Pick the other two and find $\pi \approx 3$.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §5.8 Numerical Integration, #12