Approximation of an elliptic integral

Question

I used some nice guesswork to get this formula
$$\int_0^{\frac{\pi}{2}}\sqrt{a^2\sin^2x+b^2\cos^2x}dx=\frac{ab\pi}{\left(a+b\right)\sin\left(\frac{a\pi}{a+b}\right)}$$
The comparision for some values of a and b are
$a=a,b=0, value=a, exact=a$
$a=1,b=5, value=5.236, exact= 5.2525$
$a=b, value=\frac{\pi.a}{2}, exact=\frac{\pi.a}{2} $
You can also check it gives consistent result. Am i right?

https://www.desmos.com/calculator/g0w1nkeotm

Answer 1

You are facing elliptic integrals $$I=\int_0^{\frac{\pi}{2}}\sqrt{a^2\sin^2x+b^2\cos^2x}dx=\sqrt{b^2}\, E\left(1-\frac{a^2}{b^2}\right)$$ Let us use $a=\lambda b$ and assume $b >0$ to get, as an exact result$$\frac I b= E\left(1-\lambda ^2\right)$$ that you want to compare with your rhs $$\frac{ \lambda \pi }{\lambda +1}\csc \left(\frac{ \lambda \pi}{\lambda +1}\right)$$ which seems to be "quite" good for $0 < \lambda < 2$ (look at the table below).

For the fun of it, plot both functions on the same graph. $$\left( \begin{array}{ccc} \lambda & \text{exact} & \text{approximation} \\ 0.0 & 1.00000 & 1.00000 \\ 0.1 & 1.01372 & 1.01599 \\ 0.2 & 1.04720 & 1.05050 \\ 0.3 & 1.09329 & 1.09648 \\ 0.4 & 1.14807 & 1.15066 \\ 0.5 & 1.20920 & 1.21106 \\ 0.6 & 1.27516 & 1.27635 \\ 0.7 & 1.34494 & 1.34559 \\ 0.8 & 1.41780 & 1.41808 \\ 0.9 & 1.49322 & 1.49329 \\ 1.0 & 1.57080 & 1.57080 \\ 1.1 & 1.65021 & 1.65027 \\ 1.2 & 1.73122 & 1.73145 \\ 1.3 & 1.81362 & 1.81411 \\ 1.4 & 1.89724 & 1.89807 \\ 1.5 & 1.98196 & 1.98318 \\ 1.6 & 2.06765 & 2.06931 \\ 1.7 & 2.15422 & 2.15636 \\ 1.8 & 2.24158 & 2.24423 \\ 1.9 & 2.32966 & 2.33284 \\ 2.0 & 2.41840 & 2.42211 \end{array} \right)$$

Edit

Consider $$f=E\left(1-\lambda ^2\right)\qquad \text{and} \qquad g=\frac{ \lambda \pi }{\lambda +1}\csc \left(\frac{ \lambda \pi}{\lambda +1}\right)$$ Perform Taylor expansions around $\lambda=1$ and get $$f=\frac{\pi }{2}+\frac{\pi}{4} (\lambda -1)+\frac{\pi}{32} (\lambda -1)^2-\frac{\pi}{64} (\lambda -1)^3+O\left((\lambda -1)^4\right)$$ $$g=\frac{\pi }{2}+\frac{\pi}{4} (\lambda -1)+\frac{\pi \left(\pi ^2-8\right)}{64} (\lambda -1)^2-\frac{\pi \left(\pi ^2-8\right)}{128} (\lambda -1)^3+O\left((\lambda -1)^4\right)$$ making $$f-g=\frac{\pi \left(10-\pi ^2\right)}{64} (\lambda -1)^2-\frac{\pi \left(10-\pi ^2\right)}{128} (\lambda -1)^3+O\left((\lambda -1)^4\right)$$ and, since $\pi^2 \approx 10$, this confirms, at least around $\lambda=1$ the quality of the approximation.

Answer 2

https://www.desmos.com/calculator/g0w1nkeotm

It's a pretty good approximation