Logs and sines and squares!

First, notice that using logarithm property we get $(A) \int_0^{\frac{\pi}{2}} \sin^2{x} \log^2 {\sin^2{x}} dx = 4 \int_0^{\frac{\pi}{2}} \sin^2{x} \log^2 {\sin{x}} dx$ Then, integrate by parts to get $(B) \int_0^{\frac{\pi}{2}} \sin^2{x} \log^2 {\sin{x}} dx = \int_0^{\frac{\pi}{2}} \cos^2{x} \log^2 {\sin{x}} dx + 2 \int_0^{\frac{\pi}{2}} \cos^2{x} \log {\sin{x}} dx$ We are going to compute the latter integral: using $2 \cos^2{x} = \cos{2x}+1$ we get that $(C) \int_0^{\frac{\pi}{2}} 2 \cos^2{x} \log {\sin{x}} dx = \int_0^{\frac{\pi}{2}} \cos{2x} \log {\sin{x}} dx + \int_0^{\frac{\pi}{2}} \log {\sin{x}} dx$ The last one is well known: $\int_0^{\frac{\pi}{2}} \log {\sin{x}} dx = - \frac {\pi \log{2}}{2}$

Next, we use this identity (obtained by integration by parts of the left integral again): $(D) \int_0^{\frac{\pi}{2}} \cos^2{x} \log {\sin{x}} dx = \int_0^{\frac{\pi}{2}} \sin^2{x} \log {\sin{x}} dx - \int_0^{\frac{\pi}{2}} \cos^2{x} dx$

Using the formula $\cos{2x} = \cos^2{x}-\sin^2{x}$ and that the last integral is equal to $\frac{\pi}{4}$ we plug in to compute the integral from (C): $(C) \int_0^{\frac{\pi}{2}} 2 \cos^2{x} \log {\sin{x}} dx = -\frac{\pi}{4} - \frac {\pi \log{2}}{2}$

Denote $I_{1} = \int_0^{\frac{\pi}{2}} \sin^2{x} \log^2 {\sin{x}} dx$ , $I_{2} = \int_0^{\frac{\pi}{2}} \cos^2{x} \log^2 {\sin{x}} dx$

We then get the system of equations. The integral from 1st equation is computed here or here . The second equation follows directly from (B).

$I_{1} + I_{2} = \int_0^{\frac{\pi}{2}} \log^2 {\sin{x}} dx = \frac{\pi^3}{24} + \frac{\pi \log^2{2}}{2}$

$I_{1} - I_{2} = - \frac{\pi}{4} - \frac {\pi \log{2}}{2}$

So, we get that $(A) \int_0^{\frac{\pi}{2}} \sin^2{x} \log^2 {\sin^2{x}} dx =4I_{1} = 2(I_{1} + I_{2}) + 2(I_{1} - I_{2}) = \pi \log^2 {2} - \pi \log {2} + \frac{\pi^3}{12} - \frac{\pi}{2}$

Finally, the answer is $1 \times 2 + 2 \times 1 + 2 \times 3 + 12 \times 2 = 2 + 2 + 6 + 24 = \boxed{34}$