moving limit through the integral sign Let $F(\alpha) = \int\limits_0^{\pi/2} \ln( \alpha^2 - \sin^2 x) \mathrm{d} x $ where $\alpha>1$
Im tempting to argue that $F(1) = \int\limits_0^{\pi/2} \ln (1 - \sin^2 x) dx = \int\limits_0^{\pi/2} \ln \cos^2 x dx $
But, $\alpha > 1$. Thus, the only way we can do  this is if we can do the following
$$ \lim_{\alpha \to 1^+ } \int\limits_0^{\pi/2} \ln( \alpha^2 - \sin^2 x) \mathrm{d} x = \int\limits_0^{\pi/2} \lim_{\alpha \to 1^+} \ln( \alpha^2 - \sin^2 x) \mathrm{d} x $$
Qs: are we allowed to move the limit in such a way?

It’s $\enspace\lim\limits_{\alpha\to 1^+} \ln(\alpha^2-\sin^2 x) =\ln\cos^2 x\enspace$ and we have to proof with
$$\lim\limits_{\alpha\to 1^+}((  \int\limits_0^{t\pi/2}\ln(\alpha^2-\sin^2 x)dx - \int\limits_0^{t\pi/2}\ln(\cos^2 x) dx )|_{t\to 1^-}) =\frac{\pi}{2} \lim\limits_{\alpha\to 1^+} \int\limits_0^1 \ln(1+\frac{\alpha^2-1}{\cos^2(\frac{\pi}{2}x)}) dx$$ if 
$$\lim\limits_{\alpha\to 1^+} \int\limits_0^1 \ln(1+\frac{\alpha^2-1}{\cos^2(\frac{\pi}{2}x)}) dx = \int\limits_0^1 \lim\limits_{\alpha\to 1^+} \ln(1+\frac{\alpha^2-1}{\cos^2(\frac{\pi}{2}x)}) dx =0$$ holds. 
It's $\enspace\cos(\frac{\pi}{2}x)>1-x\enspace$ for $\enspace0<x<1\enspace$ and therefore $$0< \int\limits_0^t \ln(1+\frac{\alpha^2-1}{\cos^2(\frac{\pi}{2}x)}) dx <\int\limits_0^t \ln(1+\frac{\alpha^2-1}{(1-x)^2}) dx $$ gives us an upper bound for $\enspace 0<t\leq 1 $ .  
With $\enspace a:=\sqrt{\alpha^2-1}\to 0\enspace $ for $\enspace \alpha\to 1^+\enspace $ it's
$$\int\limits_0^1 \ln(1+\frac{a^2}{(1-x)^2}) dx =$$ $$=[2a\arctan\frac{a}{1-x}-(1-x)(\ln((1-x)^2+a^2)-2)+(1-x)(\ln((1-x)^2)-2)]_0^1$$ $$\enspace \enspace = a\pi-2a\arctan a+\ln(1+a^2)\leq a\pi$$ for all $\enspace a\geq 0\enspace $ 
and it follows $$\lim\limits_{a\to 0} \int\limits_0^1 \ln(1+\frac{a^2}{(1-x)^2}) dx=0 $$ and therefore the answer to your question is Yes.
