log sin and log cos integral, maybe relate to fourier series I try to use the method of differentiation under integral sign for the first one
And integrate it back, but I failed to find the constant $c$ ....
Anyone hav other method? 
$$\begin{align}
  & \int_{0}^{\frac{\pi }{2}}{{{x}^{2}}{{\ln }^{2}}\left( 2\cos x \right)\text{d}x} \\ 
 & \int_{0}^{\frac{\pi }{3}}{x{{\ln }^{2}}\left( 2\sin \frac{x}{2} \right)}\text{d}x \\ 
\end{align}$$
 A: From
\begin{equation}
\int_{0}^{\frac{\pi }{2}}\left( 2\cos \theta \right) ^{x}\cos y\theta
d\theta =\frac{\pi }{2}F(1+\frac{x+y}{2},1+\frac{x-y}{2}),  \tag*{(1)}
\end{equation}
where $F(x,y)=\frac{\Gamma (x+y-1)}{\Gamma (x)\Gamma (y)},$ we can get that
\begin{equation}
\begin{array}{c}
\int_{0}^{\frac{\pi }{2}}\theta ^{q}\left( 2\cos \theta \right) ^{x}\cos 
\frac{2y\theta +q\pi }{2}\ln ^{p}\left( 2\cos \theta \right) d\theta  \\ 
=\frac{\pi }{2^{p+q+1}}\sum\limits_{k=0}^{q}(-1)^{q-k}C_{q}^{k}\sum
\limits_{j=0}^{p}C_{p}^{j}F_{k+j,q+p-k-j}(1+\frac{x+y}{2},1+\frac{x-y}{2})
\end{array}
\tag*{(2)}
\end{equation}
In (2) let $x,y=0,q$ replaced by $2q$ we have
\begin{equation}
\int_{0}^{\frac{\pi }{2}}\theta ^{2q}\ln ^{p}\left( 2\cos \theta \right)
d\theta =\frac{\pi }{2^{p+q+1}}\sum\limits_{k=0}^{2q}(-1)^{q-k}C_{2q}^{k}
\sum\limits_{j=0}^{p}C_{p}^{j}F_{k+j,2q+p-k-j}(1,1)  \tag*{(3)}
\end{equation}
For $F_{p,qj}(1,1)$ there is the following recurrence relations
\begin{equation}
F_{p,q}(1,1)=p!(q-1)!\sum\limits_{k=0}^{p-1}\sum
\limits_{j=0}^{q-1}C_{p+q-1-k-j}^{p-k}\frac{(-1)^{p+q-k-j}\zeta (p+q-k-j)
}{k!j!}F_{k,j}(1,1).  \tag*{(4)}
\end{equation}
By (4) we can get that
\begin{equation}
\begin{array}{c}
F_{0,0}(1,1)=1,F_{0,i}(1,1)=0,i=1,2,3,4,F_{1,1}(1,1)=\frac{\pi ^{2}}{6}, \\ 
F_{1,2}(1,1)=-2\zeta (3),F_{1,3}(1,1)=\frac{\pi ^{4}}{15},F_{1,4}(1,1)=-24
\zeta (5), \\ 
F_{2,2}(1,1)=\frac{\pi ^{4}}{90},F_{2,3}(1,1)=-2\pi ^{2}\zeta (3)-24\zeta
(5), \\ 
F_{2,4}(1,1)=\frac{68\pi ^{6}}{315}+24\zeta ^{2}(3),F_{3,3}(1,1)=\frac{
107\pi ^{6}}{420}+36\zeta ^{2}(3), \\ 
F_{3,4}(1,1)=-6\pi ^{4}\zeta (3)-48\pi ^{2}\zeta (5)-720\zeta (7), \\ 
F_{4,4}(1,1)=\frac{3701\pi ^{8}}{3150}+96\pi ^{2}\zeta ^{2}(3)-2304\zeta
(3)\zeta (5),
\end{array}
\tag*{(5)}
\end{equation}
By (3) and (5) we have
\begin{equation}
\int_{0}^{\frac{\pi }{2}}\theta ^{2}\ln ^{2}\left( 2\cos \theta \right)
d\theta =\frac{11\pi ^{5}}{1440}  \tag*{(6)}
\end{equation}
A: A related problem. One can manage to get the following  closed form solution for the second integral,
$$\int_{0}^{\frac{\pi }{3}}{x{{\ln }^{2}}\left( 2\sin \frac{x}{2} \right)}\text{d}x = \frac{1}{4}\,{_5F_4\left(1,1,1,1,1;\,\frac{3}{2},2,2,2;\,\frac{1}{4}\right)}\sim 0.2555485412 $$
