Who's up to the challenge? 65

Let $\displaystyle I = \int_{0}^{\frac{\pi}{2}} x^2 \csc x dx$

Applying IBP ,

$\displaystyle I = -2\int_{0}^{\frac{\pi}{2}} x\ln|cosecx-cotx|dx = \int_{0}^{\frac{\pi}{2}} -2x[\ln(2\sin^2\frac{x}{2}) - \ln(sinx)]dx$

Splitting into parts we get ,

$\displaystyle I = -2\int_{0}^{\frac{\pi}{2}} x\ln(2\sin^2\frac{x}{2}) dx + 2\int_{0}^{\frac{\pi}{2}} x\ln(\sin x) dx$

Both the integrals can be calculated separately and I have done that in previous cases so I'm skipping the part,

Consider a function $\displaystyle F(a,b) = \int_{0}^{\frac{\pi}{2}} \sin^a x \cos bx dx = \frac{2^{-a}\Gamma(a+1)\pi \cos(\frac{b\pi}{2})}{\Gamma(\frac{a+b+2}{2})\Gamma(\frac{a-b+2}{2})}$

We can calculate $\displaystyle \frac{\partial^2 F(1,1)}{\partial a \partial b}$ which equals our integrals .

After calculation , $\displaystyle \begin{cases} 2\int_{0}^{\frac{\pi}{2}} x\ln(2\sin^2\frac{x}{2}) dx = -2\pi G + \frac{35}{8}\zeta(3) - \frac{\pi^2}{4}\ln 2 \\ 2\int_{0}^{\frac{\pi}{2}} x\ln(\sin x) dx = \frac{7}{8}\zeta(3)-\frac{\pi^2}{4}\ln 2 \end{cases}$

So $\displaystyle I = 2\pi G-\frac{7}{2}\zeta(3)$ , Thus answer is $\boxed{2+7+2+3=14}$