Evaluating $\int_{0}^{\pi/2}\log\left(a^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)\right)\,{\rm d}x$

Note that: $${a}^{2} \cos^{2} \left( x \right) +{b}^{2} \sin^{2} \left( x \right) =\frac{1}{4} \left( 1+{\frac { \left( a-b \right) {{\rm e}^{-2\,ix}}}{a+b}} \right) \left( {\frac {{{\rm e}^{2 \,ix}} \left( a-b \right) }{a+b}}+1 \right) \left( a+b \right) ^{2}$$ $$\left|{\frac { \left( a-b\right)}{a+b}}\right|\le 1,\quad a,b>0$$ then from: $$-\sum _{n=1}^{\infty }{\frac {{r}^{n}{{\rm e}^{inx}}}{n}}=\ln \left( 1-r{{\rm e}^{ix}} \right) $$ show that: $$2\ln \left(\frac{a+b}{2} \right) -2\sum _{n=1}^{\infty } \frac{\left( -{\frac {a-b}{a+b}} \right) ^{n}\cos \left( 2\,nx \right)}{n} = \ln \left({a}^{2} \cos^{2} \left( x \right) +{b}^{2} \sin^{2} \left( x \right)\right) $$ and note that: $$\int _{0}^{1/2\,\pi }\!\cos \left( 2\,nx \right) {dx}=0,\quad n\ge 1$$ $$\int _{0}^{1/2\,\pi }\!2\ln \left( \frac{a+b}{2} \right) {dx}=\ln \left(\frac{a+b}{2} \right) \pi $$ and the integration result follows. Furthermore:

$${\frac {\partial }{\partial t}}\ln \left({t}^{2} \cos^{2} \left( x \right) +{b}^{2} \sin^{2} \left( x \right)\right) ={\frac {2t}{{t}^{2}+{b}^{2} \tan^{2} \left( x \right)}}<\frac{2}{t}$$

$$I(a)=\int_0^\frac\pi2\log\bigg(a^2\cos^2x+b^2\sin^2x\bigg)dx\quad=>\quad I'(a)=\int_0^\frac\pi2\frac{2a\cdot\cos^2x}{a^2\cos^2x+b^2\sin^2x}dx$$

$$J(b)=\int_0^\frac\pi2\log\bigg(a^2\cos^2x+b^2\sin^2x\bigg)dx\quad=>\quad J'(b)=\int_0^\frac\pi2\frac{2b\cdot\sin^2x}{a^2\cos^2x+b^2\sin^2x}dx$$

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$$I'(a)=2a\underbrace{\int_0^\frac\pi2\frac{dx}{a^2+b^2\tan^2x}dx}_{t=\tan x}=2a\,\bigg[~\frac{x-\dfrac ba\arctan\bigg(\dfrac ba\tan x\bigg)}{a^2-b^2}~\bigg]_0^\frac\pi2=\frac\pi{a+b}$$

$$J'(b)=2b\underbrace{\int_0^\frac\pi2\frac{dx}{a^2\cot^2x+b^2}dx}_{t=\cot x}=2b\,\bigg[~\frac{\dfrac\pi2-x-\dfrac ab\arctan\bigg(\dfrac ab\cot x\bigg)}{a^2-b^2}~\bigg]_0^\frac\pi2=\frac\pi{a+b}$$

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$\begin{align}F(a,b)=I(a)=\pi\ln(a+b)+C_b\\\\F(a,b)=J(b)=\pi\ln(a+b)+C_a\end{align}=>$ $F(1,1)=0=>C_a=C_b=-\pi\ln2=>F=\pi\,\ln\dfrac{a+b}2$