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
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 $2$.
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.
