Evaluating $\int_0^{\pi/2} \log \left| \sin^2 x - a \right|$ where $a\in [0,1]$. 
How to evaluate
  $$
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
$$
  where $a\in[0,1]$ ?

I think of this problem as a generalization of the following proposition
$$
\displaystyle\int_0^{\pi/2} \log  \left(\sin x\right)\,dx =-\frac12\pi\log2
$$
My try
Put
$$
I(a)=\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
$$
From the substitution $x \to \frac{\pi}{2}-x$ , we get
$$
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
=
\displaystyle\int_0^{\pi/2} \log \left| \cos^2 x - a \right|\,dx
$$
Thus
$$
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
=
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - (1-a) \right|\,dx
$$
which means $$I(a)=I(1-a) \tag{1}$$
On the other hand,
\begin{align}
2I(a) &=
\displaystyle\int_0^{\pi/2} \log \left| (\sin^2 x - a)(\cos^2 x -a) \right|\,dx
\\ &=
\displaystyle\int_0^{\pi/2} \log \left| a^2-a+\sin^2 x \cos^2 x \right|\,dx
\\ &=
\displaystyle\int_0^{\pi/2} \log \left| 4(a^2-a)+\sin^2 (2x) \right|\,dx
-\pi \log 2
\\ &=
\frac{1}{2}\displaystyle\int_0^{\pi} \log \left| 4(a^2-a)+\sin^2 x \right|\,dx
-\pi \log 2
\\ &=
\displaystyle\int_0^{\pi/2} \log \left| 4(a^2-a)+\sin^2 x \right|\,dx
-\pi \log 2
\\ &=
\displaystyle\int_0^{\pi/2} \log \left| 1+4(a^2-a)-\sin^2 x \right|\,dx
-\pi \log 2
\\ &=
I((2a-1)^2)
-\pi \log 2
\end{align}
Thus
$$
2I(a)=I((2a-1)^2)-\pi \log 2 \tag{2}
$$
Let $a=0$ we get the proposition mentioned above $\displaystyle\int_0^{\pi/2} \log  \left(\sin x\right)\,dx =-\frac12\pi\log2.$
But how to move on ? Can we solve the problem only by $(1)$ and $(2)$? Or what other properties should we use to evaluate that?
Looking forward to your new solutions as well.
Thank you in advance!
Added:
As pointed out in the comments, it seems like that the integral is identical to $-\pi\log 2$.
From $(1)$ and $(2)$ we can also find many numbers such that $I(a)=-\pi\log 2$.
 A: Using the symmetries and developing the $\sin^2$ term, we can express
\begin{align}
I&=\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left| \sin^2 x - a \right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1-2a}{2}-\frac{1}{2}\cos 2x\right|\,dx
%&=-\frac{\pi}{2}\ln 2+\frac{1}{4}\int_0^{2\pi} \log \left| \left( 2a-1 \right)+\cos 2x\right|\,dx
\end{align}
By denoting $2a-1=\cos 2\alpha$, 
\begin{align}
I&=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1}{2}\left( \cos 2\alpha+\cos 2x\right)\right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left|  \cos \left( x+\alpha \right)\cos \left( x-\alpha \right)\right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left|  \cos \left( x+\alpha \right)\right|\,dx+\frac{1}{4}\int_0^{2\pi} \log \left|  \cos \left( x-\alpha \right)\right|\,dx
\end{align} 
As the functions are periodic, the integration variables can be shifted, thus
\begin{equation}
I=\frac{1}{2}\int_0^{2\pi} \log \left|  \cos \left( x \right)\right|\,dx
\end{equation} 
Finally using the symmetries of the integrand, 
\begin{align}
I&=2\int_0^{\pi/2} \log \left|  \cos \left( x \right)\right|\,dx\\
&=2\int_0^{\pi/2} \log \left|  \sin \left( x \right)\right|\,dx\\
&=-\pi\ln 2
\end{align}
from the quoted result.
