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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? $$\begin{align} & \int_{0}^{\frac{\pi }{2}}{{{x}^{2}}{{\ln }^{2}}\left( 2\cos x \right)\text{d}x} \\ & \int_{0}^{\frac{\pi }{3}}{x{{\ln }^{2}}\left( 2\sin \frac{x}{2} \right)}\text{d}x \\ \end{align}$$

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Do you have any reason to think these have closed-form expressions? – Robert Israel Jan 29 '13 at 8:07
@RobertIsrael I think these are likely to hav one – Ryan Jan 30 '13 at 2:20
Do you have any reason for that thought? – Robert Israel Jan 30 '13 at 7:05
up vote 4 down vote accepted

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}

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A related problem. One can manage to get the following closed form solution for the second integral,

$$\int_{0}^{\frac{\pi }{3}}{x{{\ln }^{2}}\left( 2\sin \frac{x}{2} \right)}\text{d}x = \frac{1}{4}\,{_5F_4\left(1,1,1,1,1;\,\frac{3}{2},2,2,2;\,\frac{1}{4}\right)}\sim 0.2555485412 $$

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