Evaluate $\int_{0}^{\frac{\pi}{2}} \theta \log^2(\cos\theta) d\theta$

Question

$$\int_{0}^{\frac{\pi}{2}} \theta \log^2(\cos\theta) d\theta$$

$$\log(\sin(\theta)) \cdot \log(\cos(\theta)) = \frac{1}{4} \zeta(3) - \log(2) - \sum_{n=1}^{\infty} f(n) \cos(2n\theta)$$ and then we have

$$\int_{0}^{\frac{\pi}{2}} \left(\frac{1}{4} \zeta(3) - \log^3(2)\right) \theta , d\theta - \int_{0}^{\frac{\pi}{2}} \sum_{n=1}^{\infty} f(n) \theta \cos(2n\theta) , d\theta$$

$$= \frac{3}{16} \zeta(2) \zeta(3) - \frac{3}{4} \log^3(2) \zeta(2) - \sum_{n=1}^{\infty} f(n) \int_{0}^{\frac{\pi}{2}} \theta \cos(2n\theta) , d\theta$$

$$= \frac{3}{16} \zeta(2) \zeta(3) - \frac{3}{4} \log^3(2) \zeta(2) + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1 + (-1)^{n-1}}{n^2} f(n)$$

Take the series amd i expanded the integral of $f(n)$

\begin{align*} &= \frac{3}{16} \zeta(2) \zeta(3) - \frac{3}{4} \log^3 (2) \zeta(2) \ &+ \frac{1}{4} \log^2 (2) \int_{0}^{1} \frac{\text{Li}2(-t)}{t} , dt - \frac{1}{4} \log^2 (2) \int{0}^{1}\frac{\text{Li}_2(t)}{t} , dt \end{align*}

Answer 1

Utilize $$\int_0^{\pi/2}\theta \ln^2(\sin \theta)d \theta = \text{Li}_4(\frac12) -\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42 $$ and the substitution $ \theta\to \frac\pi2-\theta $ to integrate

\begin{align} &\int_0^{\pi/2} \theta\ln^2(\cos \theta)d \theta\\ =& \ \frac\pi2 \int_0^{\pi/2}\ln^2(\sin \theta)d \theta - \int_0^{\pi/2} \theta\ln^2(\sin \theta)d \theta\\ =& \ \frac\pi2 \left(\frac{\pi^3}{24}+\frac\pi2\ln^22 \right) - \left( \text{Li}_4(\frac12) -\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42\right)\\ = & -\text{Li}_4(\frac12) +\frac{79\pi^4}{2880}+\frac{\pi^2}{6}\ln^22-\frac1{24}\ln^42 \end{align}