Evaluate the integral from 0 to pi/2 of root(1 + tan^2(0.8) sin^2(beta)) with respect to beta.

Step-by-step solution

1. 1

   Identify the standard form. The complete elliptic integral of the second kind is

   $E(m)=\int_{0}^{\pi/2}\sqrt{1-m\sin^{2}\theta}\;d\theta$

   Matching $\sqrt{1-m\sin^{2}\beta}$ against $\sqrt{1+\tan^{2}(0.8)\sin^{2}\beta}$ requires $-m=\tan^{2}(0.8)$, so $m=-\tan^{2}(0.8)$ — a negative parameter, which is perfectly legal and simply means the integrand exceeds $1$ rather than falling below it.

2. 2

   Compute the parameter.

   $\tan(0.8)\approx1.02963856,\qquad\tan^{2}(0.8)\approx1.06015556$

   so $m\approx-1.06015556$ and $I=E\left(-1.06015556\right)$. Note this integral has no elementary closed form — that is exactly why it carries its own name.

3. 3

   Convert to a positive parameter (optional but clarifying). Factoring the constant out,

   $\sqrt{1+k^{2}\sin^{2}\beta}=\sqrt{1+k^{2}}\,\sqrt{1-\frac{k^{2}}{1+k^{2}}\cos^{2}\beta}$

   with $k^{2}=\tan^{2}(0.8)$, giving $I=\sqrt{1+k^{2}}\;E\!\left(\tfrac{k^{2}}{1+k^{2}}\right)$ where now $\tfrac{k^{2}}{1+k^{2}}\approx0.51460$ lies in $[0,1]$ and $\sqrt{1+k^{2}}\approx1.435324$.

4. 4

   Bracket the answer before computing it. The integrand runs from $1$ at $\beta=0$ up to $\sqrt{1+k^{2}}\approx1.4353$ at $\beta=\tfrac\pi2$. Multiplying by the interval length $\tfrac\pi2\approx1.5708$ bounds the integral between $1.5708$ and $2.2546$ — so any answer outside that range is wrong on sight.

5. 5

   Evaluate numerically. Three independent methods — a $4\times10^{6}$-node trapezoid rule, $200$-node Gauss–Legendre quadrature, and Gauss–Legendre applied to the transformed positive-parameter form — all agree:

   $I\approx1.9280090$

   This sits comfortably inside the bracket from the previous step ✓. Be careful with published values here: a nearby but incorrect figure of $1.92274$ differs in the third decimal, which is well outside the agreement of these three methods.