The value of the integral \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\log \frac{\pi-x}{\pi+x}\right) \cos x d x\) is equal to

Let \(\mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left[x^2+\log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x \mathrm{~d} x\)
\(\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \mathrm{~d} x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \mathrm{~d} x\)
Let \(\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2\)
Where \(\mathrm{I}_1=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \mathrm{~d} x\) and
\(\mathrm{I}_2=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \mathrm{~d} x\)
Consider
\(\begin{aligned}
& I_1=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \cdot d x \\
& =2 \int_0^{\frac{\pi}{2}} x^2 \cos x d x
\end{aligned}\)
\(\ldots\left[x^2 \cos x\right.\) is an even function \(]\)
\(=2\left[x^2 \cdot \int \cos x \mathrm{~d} x-\int \frac{\mathrm{d}}{\mathrm{d} x}\left(x^2\right)\left(\int \cos x \mathrm{~d} x\right) \mathrm{d} x\right]_0^{\frac{\pi}{2}}\)
\(\begin{aligned} & =2\left[x^2 \cdot \sin x-\int 2 x \cdot \sin x d x\right]_0^{\frac{\pi}{2}} \\ & =2\left[x^2 \sin x-2 \int x \cdot \sin x d x\right]_0^{\frac{\pi}{2}}\end{aligned}\)
\(=2\left[x^2 \sin x-2\left(x(-\cos x)-\int(-\cos x) \mathrm{d} x\right)\right]_0^{\frac{\pi}{2}}\)
\(\begin{aligned} & =2\left[x^2 \sin x+2 x \cos x-2 \sin x\right]_0^{\frac{\pi}{2}} \\ & =2\left[\frac{\pi^2}{4} \sin \frac{\pi}{2}+2 \times \frac{\pi}{2} \cos \frac{\pi}{2}-2 \sin \frac{\pi}{2}\right.\end{aligned}\)
\(\left.-0^2 \sin 0-2 \times 0 \times \cos 0+2 \sin 0\right]\)
\(\begin{aligned}
& =2\left[\frac{\pi^2}{4}-2-0-0-0\right] \\
& =2\left[\frac{\pi^2}{4}-2\right]=\frac{\pi^2}{2}-4
\end{aligned}\)
Consider
\(\begin{aligned}
& \mathrm{I}_2=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \mathrm{~d} x \\
& \therefore \mathrm{I}_2=0 \ldots\left[\log \left(\frac{\pi-x}{\pi+x}\right) \cos x \text { is an odd function }\right] \\
& \therefore \mathrm{I}=\mathrm{I}_1+\mathrm{I}_2 \\
& \Rightarrow I=\frac{\pi^2}{2}-4
\end{aligned}\)