calculus - Evaluating $int_0^{pi/2} log left| sin^2 x - a right|$ where $ain [0,1]$.

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 $I(a)=-\pi\log 2$.

Answer

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.