Calculate the elliptic integral integral from 0 to (pi/2)
Calculate the elliptic integral $\displaystyle\int_0^{\pi/2}\sqrt{1-\tfrac12\sin^2x}\,dx$ using Simpson's method with accuracy $0.001$.
Explanation of Method
This is the complete elliptic integral of the second kind $E(k)$ with $k^2=1/2$, which has no elementary closed form. Bound $|f^{(4)}|$ on $[0,\pi/2]$, use it in Simpson's error formula to find a sufficient subinterval count, and evaluate.
Step 1: Identify the integral
\[I=E\!\left(\frac{1}{\sqrt2}\right)=\int_0^{\pi/2}\sqrt{1-\frac12\sin^2x}\,dx.\]
Step 2: Choose a subinterval count
Since the integrand and its derivatives stay modest in size on $[0,\pi/2]$, a conservative bound $\max|f^{(4)}|\lesssim10$ combined with the Simpson error formula gives $n\ge3$; use $2n=6$ subintervals, $h=\pi/12$.
Step 3: Function values at the six nodes ($15^\circ$ steps)
| $x_i$ | $0^\circ$ | $15^\circ$ | $30^\circ$ | $45^\circ$ | $60^\circ$ | $75^\circ$ | $90^\circ$ |
| $f(x_i)$ | 1.000000 | 0.983111 | 0.935414 | 0.866025 | 0.790569 | 0.730406 | 0.707107 |
| weight | 1 | 4 | 2 | 4 | 2 | 4 | 1 |
Weighted sum $S\approx15.477244$:
\[I_{\text{Simpson}}=\frac{\pi/12}{3}\times15.477244=\frac{\pi}{36}\times15.477244\approx1.350644.\]
\[\boxed{I_{\text{Simpson}}\approx1.3506}\]
Step 4: Accuracy check
The known value is $E(1/\sqrt2)\approx1.350644$, so the actual error is only about $5\times10^{-7}$ — far inside the required $0.001$ tolerance.
Step 5: Remark
Elliptic integrals like this arise naturally in ellipse arc lengths (Problems 204–205), pendulum periods, and electromagnetic field calculations; they cannot be reduced to elementary functions, so a fast-converging numerical rule such as Simpson's is the standard way to evaluate them to any desired precision.
Result
Final answer:
\[\boxed{E(1/\sqrt2)=\int_0^{\pi/2}\sqrt{1-\tfrac12\sin^2x}\,dx\approx1.3506,\quad E\sim5\times10^{-7}}\]