Evaluate the integral from 0 to pi/2 of the square root of (1 + tan^2(0.661) sin^2(b)) with respect to b.

Step-by-step solution

1. 1

   Recognise the shape of the integrand. The complete elliptic integral of the second kind is

   $E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2\theta}\;d\theta$

   Our integrand has a plus sign, so it matches the template with $k^2=-\tan^2(0.661)$ - a negative parameter, equivalently an imaginary modulus $k=i\tan(0.661)$.

2. 2

   Compute the constant once, carefully. Everything downstream depends on this number:

   $\tan(0.661)=0.7777085,\qquad \tan^2(0.661)=0.6048305$

   (The angle $0.661$ is in radians. Using degrees here is the most common way to get a wrong answer.)

3. 3

   Restate the integral with the number in place.

   $I=\int_0^{\pi/2}\sqrt{1+0.6048305\,\sin^2 b}\;db=E\!\left(-0.6048305\right)$

   where $E$ is written with the parameter $m=k^2$ convention.

4. 4

   Bracket the answer before computing it. Since $0\le\sin^2 b\le 1$, the integrand lies between $1$ and $\sqrt{1.6048305}=1.2668$. Multiplying by the interval length $\pi/2=1.5708$ gives

   $1.5708\le I\le 1.9899$

   Any numerical result outside that window is a mistake.

5. 5

   Evaluate numerically. Composite Simpson's rule on $[0,\pi/2]$ with a fine mesh converges to

   $I\approx 1.7865275$

   This sits comfortably inside the bracket $[1.5708,1.9899]$ and just above the midpoint, as expected because $\sin^2 b$ spends more of the interval near $1$ than a linear weighting would suggest.