My first problem on Brilliant

Since $\displaystyle \beta(a,b)=2\int_0^{\pi/2} (\sin x)^{2a-1}(\cos x)^{2b-1}\; dx$ , we have by differentiating twice that

$\displaystyle \int_0^{\pi/2} \ln(\sin x)\ln(\cos x) \sin(2x)\; dx = \dfrac{1}{4}\dfrac{\partial^2}{\partial a \partial b}\left(\beta(a,b)\right)|_{a=b=1}$

It can be proved that $\displaystyle \dfrac{\partial^2}{\partial a \partial b}\left(\beta(a,b)\right)|_{a=b=1} = \left(\psi(1)-\psi(2)\right)^2-\psi^{(1)}(2)=2-\dfrac{\pi^2}{6}$

Thus $\displaystyle \int_0^{\pi/2} \ln(\sin x)\ln(\cos x) \sin(2x)\; dx = \dfrac{1}{2}-\dfrac{\pi^2}{24}$ making the answer $\boxed{29}$

Note: $\displaystyle \dfrac{\partial^2}{\partial a \partial b}\left(\beta(a,b)\right) = \beta(a,b)\left[ (\psi(a)-\psi(a+b))(\psi(b)-\psi(a+b))-\psi^{(1)}(a+b)\right]$

We can write the integral as :-

$\displaystyle 2\int_{0}^{\frac{\pi}{2}} \ln\left(\cos(x)\right)\ln\left(\sin(x)\right)\sin(x)\cos(x)\,dx$

Substituting $\sin(x) = t$ we have:-

$\displaystyle 2\int_{0}^{1}\ln\left(\sqrt{1-t^{2}}\right)t\ln(t)\,dt =\int_{0}^{1}\ln\left(1-t^{2}\right)t\ln(t)\,dt$

Now we know that $\displaystyle -\ln(1-x) = \sum_{r=1}^{\infty}\frac{x^{r}}{r}$ for $-1\leq x < 1$

Hence $\displaystyle -\ln(1-x^{2}) = \sum_{r=1}^{\infty}\frac{x^{2r}}{r}$ for $-1\leq x < 1$

Hence we have :-

$\displaystyle -\sum_{r=1}^{\infty}\int_{0}^{1}\frac{t^{2r+1}\ln(t)}{r}\,dt$

Substituting $t=e^{-z}$ we have :-

$\displaystyle \sum_{r=1}^{\infty}\int_{0}^{\infty}\frac{e^{-z(2r+2)}z}{r}\,dz$

Now substitute $(2r+2)z=y$

$\displaystyle \sum_{r=1}^{\infty}\int_{0}^{\infty}\frac{e^{-y}y}{r(2r+2)^{2}}\,dy = \frac{1}{4}\sum_{r=1}^{\infty}\frac{1}{r(r+1)^{2}}$

$\displaystyle = \frac{1}{4}\sum_{r=1}^{\infty}\frac{1}{(r+1)}\left(\frac{1}{r}-\frac{1}{r+1}\right) = \frac{1}{4}\sum_{r=1}^{\infty}\left(\frac{1}{r}-\frac{1}{r+1}\right) - \frac{1}{4}\sum_{r=1}^{\infty}\frac{1}{(r+1)^{2}}$

Note here that the splitting of the sums are possible due to convergence of both the sums by Limit Comparison Test.

$\displaystyle = \frac{1}{4} - \frac{1}{4}\left(\zeta(2)-1\right) = \frac{1}{4}+\frac{1}{4} - \frac{\pi^{2}}{24} = \frac{1}{2} - \frac{\pi^{2}}{24}$