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 

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. 
