It's an even function right?

@Ashley Shamidha – Ahh, sure :)

We have $(1)\,\, I= \large \int_{-\frac \pi 2}^{\frac \pi 2} \left [ \cos(x) \left(x^2 + \ln \left( \frac{\pi-x}{\pi+x} \right)\right) \right ] \, dx$ And by making the $u=-x$ sub we found out $(2)\,\, I= \large \int_{-\frac \pi 2}^{\frac \pi 2} \left [ \cos(u) \left(u^2 + \ln \left( \frac{\pi+u}{\pi-u} \right)\right) \right ] \, du$ But here, u is just a dummy variable. We can call it whatever we want. We can rewrite the equation (2) as $(2)\,\, I= \large \int_{-\frac \pi 2}^{\frac \pi 2} \left [ \cos(x) \left(x^2 + \ln \left( \frac{\pi+x}{\pi-x} \right)\right) \right ] \, dx$

Then adding (1) and (2) together we get

$2I=\large \int_{-\frac \pi 2}^{\frac \pi 2} \left [ \cos(x) \left(x^2 + \ln \left( \frac{\pi-x}{\pi+x} \right)\right) \right ] \, dx\\+ \large \int_{-\frac \pi 2}^{\frac \pi 2} \left [ \cos(x) \left(x^2 + \ln \left( \frac{\pi+x}{\pi-x} \right)\right) \right ] \, dx$ $2I= \large \int_{-\frac \pi 2}^{\frac \pi 2} \left [ \cos(x) \left(2x^2 + \ln \left( \frac{\pi-x}{\pi+x} \right)+ \ln \left( \frac{\pi+x}{\pi-x} \right)\right) \right ] \, dx$

The logs cancel out because $\ln(x)+\ln(\frac{1}{x})=\ln(x)+\ln(x^{-1})=\ln(x)-\ln(x)=0$