Integral $\int^{\frac{\pi}{6}}_{0}\ln^2(2\sin \theta)d\theta$

Greetings I am trying to find a close form for: $\int^{\frac{\pi}{6}}_{0}\ln^2(2\sin \theta)d\theta$ My first thought was to let $$I(k)=\int_0^{\frac{\pi}{6}} (2sin(\theta))^k d\theta$$ So I would get the answer to the original integral if I evaluate this in terms of k, derivate two times the result and plug $k=0$. I am struggling with the last one. $$I(k)=2^k\int_0^{\frac{\pi}{6}} sin^k (\theta) d\theta=2^k \,\Im\int_0^{\frac{\pi}{6}} e^{ik\theta} d\theta= 2^k \,\Im\int_0^{\frac{\pi}{6}}\frac{d}{d\theta}\left(\frac{e^{ik\theta}}{ik}\right)d\theta=\Im2^k\left(\frac{\cos k\frac{\pi}{6}+i\sin\frac{k\pi}{6}-1}{ik} \right)=2^k\frac{{1-\cos\frac{k\pi}{6}}}{k}$$ Is this correct? If not I would appreciate some help with the integral.

See the details here from a similar question Evaluating $\int_{0}^{\pi/3}\ln^2 \left ( \sin x \right )\,dx$.

The core idea is to exploit the identity,

$${\rm Re}\log^2(1-e^{i2x})=\log^2(2\sin{x})-\left(x-\frac{\pi}{2}\right)^2.$$

Alternatively, you may exploit the Fourier series of $\log(\sin(x))$ and work with a double series which is not hard to handle with.

Fourier series of Log sine and Log cos

Additional information: It's easy to show by contour integration that

$$\int_0^1 \frac{\log^2(1-x^2)}{x}\textrm{d}x+i \int_0^{\pi/6}\log^2(1-e^{i2 x})\textrm{d}x-\int_0^1 \frac{\log^2(1-e^{i\pi/3}x^2)}{x}\textrm{d}x=0,$$ from which you may extract the desired value of the integral.

$$\color{blue}{\int_0^{\frac{\pi }{6}} \ln ^2(2 \sin (t)) \, dt}$$

Substituting $2sin(t) = k$ we have: $$\int_0^1 \frac{\ln ^2(k)}{\sqrt{4-k^2}} \, dk$$

Using identity:$$\sum _{j=0}^{\infty } \frac{\Gamma \left(j+\frac{1}{2}\right) k^j}{\sqrt{\pi } j!}=\frac{1}{\sqrt{1-k}}$$

$$\int_0^1 \frac{\ln ^2(k)}{2 \sqrt{1-\frac{k^2}{4}}} \, dk=\int_0^1 \left(\sum _{j=0}^{\infty } \frac{2^{-1-2 j} k^{2 j} \Gamma \left(\frac{1}{2}+j\right) \ln ^2(k)}{\sqrt{\pi } j!}\right) \, dk=\\\sum _{j=0}^{\infty } \int_0^1 \frac{2^{-1-2 j} k^{2 j} \Gamma \left(\frac{1}{2}+j\right) \ln ^2(k)}{\sqrt{\pi } j!} \, dk=\\\sum _{j=0}^{\infty } \frac{4^{-j} \Gamma \left(\frac{1}{2}+j\right)}{(1+2 j)^3 \sqrt{\pi } j!}=\color{blue}{\frac{7 \pi ^3}{216}}$$

Sum I calculate with CAS help.