Now this is easier right?

The answer is: $\dfrac{\pi^{2}}{8}\ln^{2}(2) +\dfrac{\pi^{4}}{96}$ The $\text{SOLUTION}$
$I =\displaystyle \int_0^{\frac{\pi}{2}} x\ln^{2}(\sin 2x)\mathbb{d}x$ Using $f(a+b-x) =f(x)$ and adding the original and transformed integral, we get: $I =\dfrac{\pi}{4} \displaystyle \int_{0}^{\frac{\pi}{2}}\ln^2(sin (2x)) dx \\ =\dfrac{\pi}{4} \displaystyle \int_{0}^{\frac{\pi}{2}}\ln^2(2\sin(x)\cos(x)) dx \\$ Expanding the square term and separating terms; we get $I =\displaystyle \int_{0}^{\frac{\pi}{2}}\left[ \ln^{2}(2) + 2\ln^{2}(\sin(x))+ 4\ln(2)\ln(\sin(x)) +2\ln(\sin(x))\ln(\cos(x)) \right]dx$ where the fact that $\sin(x) =\cos(\dfrac{\pi}{2}- x)$ is used.
Further evaluating elementary integrals, and that integral of logarithm of sine and cosine over $0$ to $\pi/2$ is $-\dfrac{\pi}{2}\ln(2)$ , we get: $I = -\dfrac{3}{8}\pi^{2}\ln^{2}(2) + \dfrac{\pi}{2}\left[ \underbrace{\displaystyle \int_{0}^{\pi/2}\ln^{2}(\sin(x))dx}_{K} +\underbrace{\displaystyle \int_{0}^{\pi/2}\ln(\sin(x))\ln(\cos(x))dx}_{L} \right]$ Now to evaluate these integrals, the main work starts: Consider the following integral $\Im(a,b) = \displaystyle \int_{0}^{\pi/2} \sin^{a}(x)\cos^{b}(x) dx = \dfrac{\Gamma(\frac{a+1}{2})\Gamma(\frac{b+1}{2})}{2\Gamma(\frac{a+b+2}{2})}$ Differentiating above partially first w.rt $a$ and then w.r.t $b$ , we get $\dfrac{\partial \Im(a,b)}{\partial a \partial b} = \displaystyle \int_{0}^{\pi/2} \sin^{a}(x)\cos^{b}(x)\ln(\sin(x))\ln(\cos(x)) dx\\ = \dfrac{1}{8}\dfrac{\Gamma(\frac{1+a}{2})\Gamma(\frac{b+1}{2})}{\Gamma(\frac{a+b+2}{2})}\left[ -\psi'(\dfrac{a+b+2}{2}) + \left[ \psi(\frac{b+1}{2}) -\psi(\frac{a+b+2}{2})\right]\left[\psi(\frac{1+a}{2}) -\psi(\frac{a+b+2}{2})\right] \right]$ where $\Gamma(x)$ is the Gamma Function and $\psi(x)$ and $\psi'(x)$ are the Digamma and Trigamma Functions Respectively.
Now, Plugging $a=0$ and $b=0$ , we get: $L =\displaystyle \int_{0}^{\pi/2} \ln(\sin(x))\ln(\cos(x)) dx = -\dfrac{\pi^3}{48} +\dfrac{\pi}{2}\ln^{2}(2)$ where, the fact that: $\Gamma(1/2) =\sqrt{\pi}$ , $\psi(1) = -\gamma$ , $\psi(1/2) =-2\ln(2) -\gamma$ and $\psi'(1) =\dfrac{\pi^{2}}{6}$ are used. Here, $\gamma$ is the Euler Mascheroni Constant.
Now $H_{n}$ is the Harmonic Number and
For evaluating $\psi'(1)$ , the following consideration was used: $\psi(n) =H_{n-1} -\gamma$ Differentiating w.r.t $n$ ; $\psi'(n) = (H_{n-1})'$ but as $H_{n-1} =\displaystyle \int_{0}^{1} \dfrac{1-x^{n-2}}{1-x}dx$ $\psi'(n$ can be calculated. Now , for calculating $K$ , the integral: $\Im(a) = \displaystyle \int_{0}^{\pi/2} \sin^{a}(x) dx$ is used and is Differentiated twice.This has been left as an exercise to the reader. The intregral comes out to be: $K =\displaystyle \int_{0}^{\pi/2}\ln^{2}(\sin(x))dx =\dfrac{\pi}{2}\ln^{2}(2) + \dfrac{\pi^{3}}{24}$ So, the final Integral Comes out to be: $\large \boxed{\displaystyle \int_0^{\frac{\pi}{2}} x\ln^{2}(\sin 2x)\mathbb{d}x = \dfrac{\pi^{2}}{8}\ln^{2}(2) +\dfrac{\pi^{4}}{96 } }$