Integral $\int_{0}^{\pi/3}\ln^4\left(\frac{\sin x}{\sin(x+\pi/3)}\right)dx$ 
Calculate$$\int_{0}^{\pi/3}\ln^4\left(\frac{\sin x}{\sin(x+\pi/3)}\right)\mathrm dx$$

My try:
$$\sin(x+\pi/3)=\sin x\cos(\pi/3)+\sin(x+\pi/3)\cos x=\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x$$
$$\int_{0}^{\pi/3}\ln^4\left(\frac{2\sin x}{\sin x+\sqrt{3}\cos x}\right)\mathrm dx$$
$$\int_{0}^{\pi/3}\ln^4\left(\frac{2}{1+\sqrt{3}\cot x}\right)\mathrm dx$$
$u=\frac{2}{1+\sqrt{3}\cot x}$
$u^{'}=\frac{2\sqrt{3}}{\sin^2 x(1+\sqrt{3}\cot x)^2}$
$$\frac{1}{2\sqrt{3}}\int_{0}^{1}\sin^2 x(1+\sqrt{3}\cot x)^2\ln^4(u)\mathrm du$$
$$\frac{2}{\sqrt{3}}\int_{0}^{1}\sin^2 x u^{-2}\ln^4(u)\mathrm du$$
$\cot^2 x=\frac{(2-u)^2}{3u^2}$
using $1+\cot^2 x=\frac{1}{\sin^2 x}$
$$\frac{2}{\sqrt{3}}\int_{0}^{1}\frac{3u^{2}}{3u^2+(2-u)^2} u^{-2}\ln^4(u)\mathrm du$$
$$\frac{6}{\sqrt{3}}\int_{0}^{1}\frac{1}{3u^2+(2-u)^2}\ln^4(u)\mathrm du$$
$$\frac{3}{\sqrt{3}}\int_{0}^{1}\frac{1}{u^2-u+1}\ln^4(u)\mathrm du$$
I am not sure what to do next...
 A: $$I=\int_{0}^{\pi/3}\ln^4\left(\frac{\sin(x+\pi/3)}{\sin x}\right)dx=\int_0^\frac{\pi}{3}\ln^4 \left(\frac12 +\frac{\sqrt 3}{2}\cot x\right)dx=\frac{\sqrt 3}{2}\int_0^1 \frac{\ln^4 x}{x^2-x+1}dx$$
Above follows by the substitution $\frac12+\frac{\sqrt{3}}{2}\cot x\to x$. We also have for $t\in(0,\pi), x\in (-1,1)$:
$$\frac{\sin t}{x^2-2x\cos t+1}=\sum_{n=1}^\infty x^{n-1}\sin(nt)$$
$$\Rightarrow I=\sum_{n=1}^\infty  \sin\left(\frac{n\pi}{3}  \right)\int_0^1 x^{n-1} \ln^4 x dx=24\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}  \right)}{n^5}$$

Similarly to here, we have for $x\in(0,2\pi)$:
$$\frac{\pi-x}{2}=\sum_{n=1}^\infty\frac{\sin(nx)}{n}$$
Integrating the above with respect to $x$ gives:
$$\sum_{n=1}^\infty \frac{\cos(nx)}{n^2}=\frac{(\pi-x)^2}{4}+C_1$$
Setting $x=\pi$ gives $C_1=-\frac{\pi^2}{12}$ and
integrating again produces:
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=-\frac{(\pi-x)^3}{12}-\frac{\pi^2}{12}x+C_2$$
Putting $x=\pi $ yields $C_2= \frac{\pi^3}{12}$. One more time:
$$\sum_{n=1}^\infty \frac{\cos(nx)}{n^4}=-\frac{(\pi-x)^4}{48}+\frac{\pi^2}{24}x^2-\frac{\pi^3}{12}x+C_3$$
For $x=\pi \Rightarrow C_3=\frac{23\pi^4}{720}$. And finally, one more similar step gives for $x \in(0,2\pi)$:
$$S(x)=\sum_{n=1}^\infty \frac{\sin(nx)}{n^5}=\frac{(\pi-x)^5}{240}+\frac{\pi^2}{72}x^3-\frac{\pi^3}{24}x^2+\frac{23\pi^4}{720}x-\frac{\pi^5}{240}$$

And well, the value of the integral is just $24S\left(\frac{\pi}{3}\right)$. Doing the algebra yields:
$$\boxed{\int_{0}^{\pi/3}\ln^4\left(\frac{\sin x}{\sin(x+\pi/3)}\right)dx=\frac{17\pi^5}{243}}$$
