We know the property

$$\begin{aligned} & \int_{-a}^a f(x) d x \\ & =\left\{\begin{array}{cc} 2 \int_0^a f(x) d x, & \text { when } f(x) \text { is an even function } \\ 0, & \text { when } f(x) \text { is an odd function } \end{array}\right. \end{aligned}$$

Let, $$\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\ln \left(\frac{\pi+x}{\pi-x}\right)\right) \cos x \cdot d x$$

$$\Rightarrow \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot \cos x d x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x \cdot \ln \left(\frac{\pi+x}{\pi-x}\right) d x$$

Here, $$x^2 \cos x$$ is an even function and $$\cos x \cdot \ln \left(\frac{\pi+x}{\pi-x}\right)$$ is an odd function.

$$\begin{aligned} & \therefore \mathrm{I}=2 \int_0^{\frac{\pi}{2}} x^2 \cos x d x+0 \\ & \Rightarrow \mathrm{I}=2\left[\left(\sin x \cdot x^2\right)_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 2 x \sin x \cdot d x\right] \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4 \int_0^{\frac{\pi}{2}} x \sin x \cdot d x \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4\left[(x(-\cos x))_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 1 \cdot(-\cos x) d x\right] \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4 \int_0^{\frac{\pi}{2}} \cos x d x \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4[\sin x]_0^{\frac{\pi}{2}} \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4 \end{aligned}$$