The value of \(I=I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \mathrm{~d} x\) is equal to

\(\begin{array}{r}
I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \mathrm{~d} x ...(i)\\
I=\frac{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\left(\frac{-\pi}{2}+\frac{\pi}{2}-x\right)^2 \cdot \cos \left(-\frac{\pi}{2}+\frac{\pi}{2}-x\right)}{1+e^{\left(-\frac{\pi}{2}+\frac{\pi}{2}+x\right)}}}{\quad \ldots\left[\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]}
\end{array}\)
\(=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+\mathrm{e}^x}\)
Adding equation (i) and (ii), we get
\(\begin{aligned} & 2 \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\mathrm{e}^x}+\frac{x^2 \cos x}{1+\mathrm{e}^{-x}}\right) \mathrm{d} x \\ & 2 \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x\left[\frac{1}{1+\mathrm{e}^x}+\frac{1}{1+\mathrm{e}^{-x}}\right] \mathrm{d} x\end{aligned}\)
\(\begin{aligned} & 2 \mathrm{I}=\int_{\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x\left[\frac{1}{1+\mathrm{e}^x}+\frac{\mathrm{e}^x}{\mathrm{e}^x+1}\right] \mathrm{d} x \\ \therefore \quad & 2 \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot \cos x \mathrm{~d} x\end{aligned}\)
\(\therefore \quad 2 \mathrm{I}=2 \int_0^{\frac{\pi}{2}} x^2 \cos x \mathrm{~d} x\)
\(\therefore\left[\begin{array}{l}\int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(x)=2 \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x \\ \text { if } \mathrm{f}(x) \text { is even function }\end{array}\right]\)
\(\begin{aligned} \therefore \quad I & =\int_0^{\frac{\pi}{2}} x^2 \cos x \cdot d x \\ & =\left[x^2 \cdot \sin x-2 \int x \cdot \sin x \mathrm{~d} x\right]_0^{\frac{\pi}{2}} \\ & =\left[x^2 \sin x-2\left(-x \cos x+\int \cos x \mathrm{~d} x\right)\right]_0^{\frac{\pi}{2}} \\ & =\left[x^2 \sin x+2 x \cos x-2 \sin x\right]_0^{\frac{\pi}{2}} \\ & =\left(\frac{\pi^2}{4}-2 \sin \frac{\pi}{2}-0+0-0\right) \\ & =\frac{\pi^2}{4}-2\end{aligned}\)