Complex integration with trigonometric and logarithm 
Show that $\int_0^{2\pi}\log\sin^22\theta dx=4\int_0^\pi\log\sin \theta d\theta=-4\pi
 \log2$

I did $$\int_0^{2\pi}\log\sin^22\theta d\theta=4\int_0^{\frac{\pi}{4}}\log\sin^22\theta d\theta$$
taking $u=2\theta$
$$2\int_0^\frac{\pi}{2}\log\sin^2\theta d\theta$$
but I need to get $\sin \theta$ and $\sin^2 \theta=\frac{1}{2}(1-\cos2\theta)$ does not help much. I tried some trigonometric manipulation but could not get anything 
 A: First note that 
$$\int_0^{2\pi}\log\left(\sin^2(2\theta)\right)\,d\theta=\frac12\int_0^{4\pi}\log \left(\sin^2 (\theta)\right) \,d\theta$$
Next, exploiting the fact that $\sin^2(x\pm n\pi)=\sin^2 (x)$ for all integer values of $n$, reveals that
$$\frac12 \int_0^{4\pi}\log\left( \sin^2(x)\right)\,dx=2\int_0^{\pi}\log\left( \sin^2(x)\right)\,dx$$
Now, using $\log (x^n)=n\log (x)$, for $x>0$ shows
$$2\int_0^{\pi}\log\left(\sin^2(\theta)\right)\,d\theta=4\int_0^{\pi}\log\left(\sin (\theta)\right)\,d\theta=-4\pi \log (2)$$
The integral on the right-hand side is [well-know] and is equal $-\pi \log (2)$ from which we obtain the expected result.  To show this we note that
$$\begin{align}
\int_0^{\pi}\log\left(\sin (\theta)\right)\,d\theta&=\int_0^{\pi/2}\log\left(\sin (\theta)\right)\,d\theta+\int_{\pi/2}^{\pi}\log\left(\sin (\theta)\right)\,d\theta \tag 1\\\\
&=\int_0^{\pi/2}\log\left(\sin (\theta)\right)\,d\theta+\int_{0}^{\pi/2}\log\left(\cos (\theta)\right)\,d\theta \tag 2\\\\
&=\int_0^{\pi/2}\log \left(\sin( x) \cos (x)\right)\,dx \tag 3\\\\
&=\int_0^{\pi/2}\log \left(\frac12 \sin (2x)\right)\,dx \tag 4\\\\
&=-\frac{\pi}{2}\log (2)+\int_0^{\pi/2}\log \left(\sin (2x)\right)\,dx \tag 5\\\\
&=-\frac{\pi}{2}\log (2)+\frac12 \int_0^{\pi}\log \left(\sin (x)\right)\,dx\tag 6\\\\
\frac12\int_0^{\pi}\log\left(\sin (\theta)\right)\,d\theta&=-\frac{\pi}{2}\log (2) \tag 7\\\\
\int_0^{\pi}\log\left(\sin (\theta)\right)\,d\theta&=-\pi \log (2) \tag 8
\end{align}$$
$(1)$ Split the integral
$(2)$ Change variables ($x \to x+\pi/2$) in the second integral and use $\sin (x+\pi/2)=\cos (x)$.
$(3)$ Combine integrals and exploit $\log (x)+\log (y)=\log (xy)$.
$(4)$ Exploit $\sin (2x)=2 \sin (x)\cos (x)$.
$(5)$ Use $\log( xy) =\log (x )+\log (y)$ and compute the integral of $\log \left(\frac12\right)$.
$(6)$ Change variables $x\to 2x$.
$(7)$ Subtract $\frac12 \int_0^{\pi/2}\log\left( \sin (x)\right) dx$ from both sides.
$(8)$ Multiply both sides by $2$.
A: Hint: $$\log \sin^2\theta=\log (\sin\theta)^2=2\log\sin\theta$$
A: Consider what has already been demonstrated
\begin{align} 
\int_{0}^{2\pi}\log(\sin^{2}(2\theta)) \, d\theta &= \frac{1}{2} \, \int_{0}^{\pi} \log(\sin^{2}(\theta)) \, d\theta = \frac{I}{2}
\end{align}
Now let $2 \sin^{2}(\theta) = 1 - \cos(2\theta)$ to obtain
\begin{align}
I &= - \ln(2) \, \int_{0}^{\pi} d\theta + \int_{0}^{\pi} \ln(1-\cos(2\theta)) \, d\theta \\
&= - \pi \, \ln 2 - \sum_{n=1}^{\infty} \frac{1}{n} \, \int_{0}^{\pi} \cos^{n}(2\theta) \, d\theta \\
&= - \pi \, \ln 2 - \sum_{n=1}^{\infty} \frac{1}{2 n} \, \int_{0}^{2\pi} \cos^{n}(x) \, dx  \hspace{5mm} x = 2\theta \\
&= - \pi \, \ln 2 - \sum_{n=1}^{\infty} \frac{1}{2 n} \, \left( \int_{0}^{\pi} \cos^{n}(x) \, dx + \int_{\pi}^{2\pi} \cos^{n}(x) \, dx \right) \\
&= - \pi \, \ln 2 - \sum_{n=1}^{\infty} \frac{1+(-1)^{n}}{2 n} \, \int_{0}^{\pi} \cos^{n}(x) \, dx \\
&= - \pi \, \ln 2 - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2n} \cdot \frac{(2)^{2} \, \Gamma\left(\frac{1}{2} \right) \, \Gamma\left(\frac{2n+3}{2}\right)}{(2n+1) \, \Gamma\left(n + 1\right)} \\
I &= - \pi \, \ln 2 - \pi \, \ln 2 = - 2 \pi \, \ln 2.
\end{align}
With this result it is evident that
\begin{align}
\int_{0}^{2\pi} \ln(\sin^{2}(2\theta)) \, d\theta = - \pi \, \ln 2.
\end{align}
This result may also be obtained by the proper changes and the link to a former question presented by Chapper's comments. 
