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Verify the following solution very carefully and thoroughly:
_______________________
The mathematical solution in the image is as follows:

```latex
text{The following is a solution in large steps.}

text{First, let's integrate by parts:}

I = int_{0}^{frac{pi}{2}} left(x^2 csc^2(x) - frac{pi^2}{4} 
ight) tan(x) ln(cos(x)) , dx 
= int_{0}^{frac{pi}{2}} left(x csc^2(x) - x^2 cot(x) csc^2(x) 
ight) ln^2(cos(x)) , dx.

text{Using the change of variables } x mapsto arctan(x) text{ and expanding yields}

I = frac{1}{4} underbrace{int_{0}^{infty} frac{arctan(x) ln^2(1+x^2)}{x^2} , dx}_{I_1} - frac{1}{4} underbrace{int_{0}^{infty} frac{arctan^2(x) ln^2(1+x^2)}{x^3} , dx}_{I_2}.

text{Now, for the resulting integrals, we will exploit the key identity:}

ln(1+iz) = frac{1}{2} ln(1+z^2) + i arctan(z), quad z in mathbb{C} land z 
eq pm i.

text{So, this means that}

I_1 = int_{0}^{infty} frac{arctan(x) ln^2(1+x^2)}{x^2} , dx 
= frac{4}{3} Im left{ int_{0}^{infty} frac{ln^3(1+ix)}{x^2} , dx 
ight} + frac{4}{3} int_{0}^{infty} frac{arctan^3(x)}{x^2} , dx 
= 8 zeta(3) + frac{4}{3} int_{0}^{infty} frac{arctan^3(x)}{x^2} , dx.

text{Furthermore, we have}

I_2 = int_{0}^{infty} frac{arctan^2(x) ln^2(1+x^2)}{x^3} , dx 
= - frac{2}{3} 
e left{ int_{0}^{infty} frac{ln^4(1+ix)}{x^3} , dx 
ight} + frac{1}{24} int_{0}^{infty} frac{ln^4(1+x^2)}{x^3} , dx + frac{2}{3} int_{0}^{infty} frac{arctan^4(x)}{x^3} , dx 
= - frac{2}{3} left( frac{2 pi^4}{15} - 12 zeta(3) 
ight) + frac{1}{24} left( frac{2 pi^4}{15} 
ight) + frac{4}{3} int_{0}^{infty} left( frac{1}{x^2} - frac{1}{1+x^2} 
ight) arctan^3(x) , dx 
= - frac{5 pi^4}{48} + 8 zeta(3) + frac{4}{3} int_{0}^{infty} frac{arctan^3(x)}{x^2} , dx.

text{This means that}

I_1 - I_2 = 8 zeta(3) + frac{4}{3} int_{0}^{infty} frac{arctan^3(x)}{x^2} , dx - left( - frac{5 pi^4}{48} + 8 zeta(3) + frac{4}{3} int_{0}^{infty} frac{arctan^3(x)}{x^2} , dx 
ight) 
= frac{5 pi^4}{48},

text{and thus}

I = frac{1}{4} (I_1 - I_2) = frac{1}{4} left( frac{5 pi^4}{48} 
ight) = frac{5 pi^4}{192}.
```