How to evaluate the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)?$ Proposed:

$$\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)\tag1$$
  Where $a\ge b$

Examples:
Where $F(1,1)= \sqrt{2}$, $F(2,1)=\sqrt{2-\sqrt{3}}$, $F(3,1)={1\over \sqrt{3}}(2-\sqrt{2})$, $...$
How do we evaluate the closed form for $(1)?$

You may like this method. Let 
$$I(b)=\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right) $$
and then
\begin{eqnarray}
I'(b)&=&\int_{0}^{\pi/2}\bigg(\frac1{a+b\sin^2(x)}+\frac1{a-b\sin^2(x)}\bigg) \mathrm dx\\
&=&\frac{\pi}{2\sqrt a}(\frac1{\sqrt{a+b}}+\frac{1}{\sqrt{a-b}}).
\end{eqnarray}
So
$$ I(b)=\frac{\pi}{2\sqrt{a}}\int_0^b(\frac1{\sqrt{a+t}}+\frac{1}{\sqrt{a-t}})dt=\pi\frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt a}$$
and hence
$$　F(a,b)=\frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt a}. $$
