Do orthodox methods always work?

Alternate Approach :

Consider the integral $\displaystyle J(a) = 2^a\int_{0}^{\frac{\pi}{2}} \sin^a(x)dx = \frac{2^{a-1}\Gamma(\frac{a+1}{2})\sqrt{\pi}}{\Gamma(\frac{a}{2}+1)}$

Our integral is equal to $\displaystyle I = \int_{0}^{\frac{\pi}{2}}(\ln(\cos^2 x))^2dx = 4J''(0) + 2\pi\ln^2 2$

On calculation we have $\displaystyle J''(0)=\frac{\pi^3}{24}$ which turns $\displaystyle I=\frac{\pi^3}{6}+2\pi\ln^2 2$ making the answer $\boxed{3+6+2+2+2=15}$

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2m-1 }\left( x \right) { cos }^{ 2n-1 }\left( x \right) dx } =B\left( m,n \right) \\ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2m-1 }\left( x \right) { \left( { cos }^{ 2 }x \right) }^{ \frac { 2n-1 }{ 2 } }dx } =B\left( m,n \right)$

On differentiating it twice w.r.t. to n and taking $m=\frac{1}{2}$ and $n=\frac{1}{2}$ , we get

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\ln { \left( { cos }^{ 2 }x \right) dx })^2 } =\frac { 1 }{ 2 } \frac { { \left( \Gamma \left( \frac { 1 }{ 2 } \right) \right) }^{ 2 } }{ 1 } \left\{ { \left( \psi \left( \frac { 1 }{ 2 } \right) -\psi \left( 1 \right) \right) }^{ 2 }+\psi '\left( \frac { 1 }{ 2 } \right) -\psi '\left( 1 \right) \right\}$

$\therefore \int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\ln { \left( { cos }^{ 2 }x \right) dx })^2 } =\frac { { \pi }^{ 3 } }{ 6 } +2\pi { \left( \ln { 2 } \right) }^{ 2 }$