Integral Integral please please

If we define $F(b) \; = \; \int_{-\frac12\pi}^{\frac12\pi}\ln(b + \sin x)\,dx \; = \; \tfrac12\int_{-\pi}^\pi \ln(b + \sin x)\,dx \; = \; \tfrac12\int_{-\pi}^\pi \ln(b + \cos x)\,dx$ for $b \ge 1$ , then $\begin{aligned} F(1) & = \; \tfrac12\int_{-\pi}^\pi \ln(2\cos^2\tfrac12x)\,dx \; = \; \int_0^\pi \ln(2\cos^2\tfrac12x)\,dx \; = \; 2\int_0^{\frac12\pi}\ln(2\cos^2x)\,dx \\ & = \; \pi\ln2 + 4\int_0^{\frac12\pi}\ln(\cos x)\,dx \; = \; \pi\ln2 + \left.\frac{\partial B}{\partial \alpha}\right|_{\alpha =\beta=\frac12} \\ & = \; \pi\ln2 + B(\tfrac12,\tfrac12)\big[\psi(\tfrac12)-\psi(1)\big] \; = \; -\pi\ln2 \end{aligned}$ while $\begin{aligned} F'(b) & = \; \tfrac12\int_{-\pi}^\pi \frac{dx}{b + \cos x} \; = \; \int_{|z|=1}\frac{1}{2b + z + z^{-1}} \frac{dz}{iz} \; = \; -i\int_{|z|=1}\frac{dz}{z^2 + 2bz + 1} \\ & = \; -i\int_{|z|=1} \frac{dz}{(z - u_+)(z - u_-)} \; = \; 2\pi\,\mathrm{Res}_{z=u_+} \frac{1}{(z-u_+)(z-u_-)} \;= \; \frac{2\pi}{u_+ - u_-} \;= \; \frac{\pi}{\sqrt{b^2-1}} \end{aligned}$ for $b > 1$ , where $u_\pm = -b \pm\sqrt{b^2-1}$ . The desired double integral is $\begin{aligned} \int_0^1 \left(\int_{-\frac12\pi}^{\frac12\pi}\ln(a^{-1} + \sin x)\,dx\right)\,da & = \; \int_0^1 F\big(a^{-1}\big)\,da \; = \; \int_1^\infty \frac{F(b)}{b^2}\,db \\ & = \; \left[-\frac{F(b)}{b}\right]_1^\infty + \int_1^\infty \frac{F'(b)}{b}\,db \; = \; \pi\int_1^\infty \frac{db}{b\sqrt{b^2-1}} + F(1) \\ & = \; \pi\int_0^1 \frac{da}{\sqrt{1-a^2}} + F(1) \; = \; \tfrac12\pi^2 + F(1) \\ & = \; \tfrac12\pi(\pi - \ln4) \end{aligned}$ making the answer $\boxed{4}$ .